WBUT Solved Question Papers EE Electromacnetism B Tech Ist Sem

WBUT Solved Question Papers EE

Electromagnetism B Tech Ist Sem

Self inductance of a magnetic coil is proportional to

[WBUT 2002]

a) N                       b)1/N                     c)N*       d)1/N*

[N = No. of turns]

Answer: (c)

. Flux in a magnetic circuit can be compared in an electric circuit to[WBUT 2006] a) voltage    b) current               c) resistance d) inductance

Answer: (c)                                                               .

For a coil with N-tums the self-inductance will be proportional to        –

a) N              b) —                c) N2 d) -Xr

N                                    N

■                                                                                                                                                                                                                                                                     –  [WBUT 2006, 2009(DECEMBER), 2011 December]

Answer: (c)

iXlh* reluctance of a magnetic circuit depends upon its                                                                         [WBUT 2007]

a) length               b) cross-section and length                       c) resistivity

Answer: (b)                                                           –

‘^Ll.5. Hysteresis loss in a transformer can be reduced by using [WBUT 2007,2008] < a) laminated Core                       . b) silicon Steel Core

c)                                                                                                                                                                                                                                oil                                                d) none of these                                                /

Answer: (a)

X1 -6. The reluctance of a magnetic circuit is given by                                                                         [WBUT 2008(JUNE)]

a)                         b) 4r, e) 1/ft Ad) ‘

JV1                               ftrA

Answer: (a)

7. The unit of magnetic flux density is                                                                         [WBUT 2009(«JUNE)]

a)                                                                                                                   weber                     b) tesla                        c) coulomb                d) none of these •

Answer: (b)

.8. The reluctance of a magnetic circuit depends on its                                                                      [WBUT 2009(JUNE)]

a)                                                                                                                                                                                                                                                                                  length                                                          b) cross-sectional area and length

c) resistivity                                                   d) cross-sectional area

Answer: (d)                                                                      .


1.9.                                                                                                                                                                                                                                                                        Which of the following is not true of leakage flux?                                                                    [WBUT 2010(JUNE)]

a)                                                                                                                                                                                                                                                                                                                             it links both the winding through air                                                                        .

b)     It links the primary winding through afr

c)      It links the secondary winding through air

d)      It does not link both the windings

Answer: (d)                    Xt

1.10. What is done to balance the mm/setup due to the secondary current?

a)    The primary voltage is increased    [WBUT 2010(JUNE)]

b)    The core flux is increased immediately

c)     The current in the primary is increased

d)     All of these

Answer: (c)                                                                                                        ■

<11. In a magnetic circuit, once a flux is set up                                                                        [WBUT 2010(JUNE)]

a)     no further energy is required

b)     energy is continuously required to maintain the flux

c)     energy Is released In the fctffti of heat                        :

d)      none of these                               .

Answer: (b)                                                                          .

. The force experienced by a small conductor of length L, carrying a current /, placed in a magnetic field B at an angle 9 with respect to B is given by

a)     BiL b) BIL sin 6 c) BIL cos 9 d) zero {WBUT 2010(DECEMBER)]

1.13. The im^ual inductance between two coupled coils is 10 mH. If turns of one

coil are doubted and tint in other are halved, the mutual inductance will be

a) 5mH b) 10mH

c)14mH                  d) 20mH [WBUT 2010( DEC EMBER)]

Answer: (b)                                   ‘                                                          .

1.14 Area hysteresis loop is a measure of                                                                      [WBUT 2011 (DECEMBER)]

a)                                                                                                                                                                                                                                                                                                retentivity                                                            b> coercivity

c)                                                                                                                                                                                                                                                                                            saturated flux density                                                d) energy loss

Answer: (d)


2Wb/m‘t (Tesla). If the conductor moves with a velocity of 5dm see1, find the induced e.m.f. when it is (i) at right angle (ii) at an angle of 30° and (iii) parallel to the magnetic field.                                                                      [WBUT 2004]


Magnitude of the induced e.m.f- when a conductor moves in a magnetic field is expressed by e = Blv sin 6

(i)           When 0 = 90°, e = 2 x 0.8 x 50 = 80 V

(ii)                 * = 2×0.8x50sin30° =40 V

(iii) When 0 = 0°, * = 2×0.8x50sin0° =0V

2.2.    An air-cored toroidal coil has 450 turns and a mean diameter of 300 mm and cross-sectional area of 300 mm2. Determine the self-inductance of the coil and the average voltage induced in it when a current of 2A is reversed in 40 ms.

{WBUT 2005]


The number of turns of the ceil N = 450

Effective length i = &d = ttx300mm = 942.478mm = 0.942m

Cross sectional area A = 500 mm2 = 300 xlO’6 m2 (i = Permeability of air = permeability of free space

= MrM0= lx4*rxl(T7 _ # – Therefore m self-inductance of the coil ■ ■

, N2/iA 450x450x4;rx300xl(r6xICr7

L~———————————————————- ————- = 8l.041{iH

£                                     0.942                                       –

The average voltage induced in an inductive coil

Tdi 81.04×10^x2 ■                                                      ‘ ’ ‘ ‘

vts=z‘-r=—————– y = 4.052 mV

• dt 40×10

2.3.  Justify your answer.

In a.c. circuit, laminated iron is invariably used in order to reduce eddy current

JesJws                   –                                                                                                   [WBUT 2005]


True,                                                                .                                                                         ■                                                                         ,

The term “eddy currents” is applied to those electric currents, which circulate within a mass of conducting material when the latter is situated in a varying magnetic field. The conducting material may be considered as consisting of a large number of closed conducting paths, eacfo of which behaves like the short-circulated winding of the transformer of which the varying magnetic field is the working flux.

The core of any electromagnetic machine is subjected to varying field fluxes. This gives rise to ‘Eddy emfs’ in elemental paths within the solid iron core facilitating the flow of eddy currents. The eddy currents since they flow in closed paths in the material – usually iron – have an axial magnetic field of their own which is in opposition to the inducing magnetic field, and so reduces the strength. These eddy currents result in heating of the core which amount to increased core losses.

If the core is laminated, the possibility of presence of closed conducting paths in the solid

core is less; thereby impeding the flow of eddy currents. This process obviously reduces eddy current losses. . ■ ■■                                                                                                  ‘                                                                                                   .

2.4.  Derive an expression for the hysteresis loss in a magnetic material.

–                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                     [WBUT 2008(DECEMBER)]


Hysteresis Loss

Flux density <B)
TThe Hysteresis Loop
Field strength
Fig. 1 shows the effect of hysteresis in ferromagnetic materials

We start with an unmagnetized sample at the origin (P,) where both field strength and

flux density are zero. The field strength is increased in the positive direction and the flux

begins to grow along the dotted path until we reach P2. This is called the initial magnetization curve.

If the field strength is now relaxed then some curious behavior occurs. Instead of retracing the initial magnetization curve the flux falls more slowly. In fact, even when the applied field is returned to zero there will still be a remaining (jremnant or remanent) flux density at P3. It is this phenomenon which makes permanent magnets possible.

To force the flux to go back to zero we have to reverse the applied field (P4). The field strength here is called the coercivity. We can then continue reversing the field to get to

P5, and so on round this type of magnetization curve called (by J. A. Ewing) a hysteresis


The expression of Hysteresis loop can be written as Wh = KUV fB16W

Where Kk is a constant which depends on material and range of flux density , V is

volume of the core material, f is the frequency of the alternation of the current passing through the magnetizing coil, Bm is the maximum value of flux density in the core in Wb/m2

2.5.    State Biot-Savart law applicable to electro-magnetism. [WBUT 2006(JUNE)] Answer!

Biot Savart Law                                                                                            –

Definition:                                                                                            – /

It was showed by Biot and Savart, through experiment that the magnetic field strength is

inversely proportional to the distance from the conductor in which the current flows. Mathematical Derivation: –                                                                                    dB

We require an expression for the B-field due to a current element. The geometry is shown in the figure. We requite the field dB at a point P a distance r from a current element of length dl (described by a vector dl). The current element carries a current I, and die vector (r) from the current element to P makes an angle 9 to dl.

Experiments show that dB is inversely proportional to ? & proportional to the length of the current element (dl) and the cuiTent (/), which flows and also sin ft Hence the field

due to the current element can be written as dB — —                                                                       –


The resultant field is foupd to be normal to the plane containing the vectors r and dl so

that the equation for the field can be written as dB

Anr\ ‘f

A m                       –

where r is a unit vector along r. This expression is known as the ‘Biot-Savart law’.

The strength of the magnetic interaction is defined by po, which is known as the permeability of free space, jio has a value of 4%x 1 O’7 Hiti1 (Henry/meter). :

The unit of B is the tesla (T)”                                •                                                                                          ^ . .*? ,

By use of a suitable integration the Biot-Savart law can be used to calculate the B-field resulting from a circuit L that can be decomposed into an infinite number of connected

current elements B = cf ~.

Applications Bio-Savart Law                                                                                          . .

Bio-Savart law can be applied to find the magnetizing force (H) produced by current currying conductor, current carrying coil of different configurations and the solenoids;

Let us consider an infinitesimal length’ dl of the wire AB and a certain point P at a distance Jt from it. According to the Bio-Savart law, the small amount of magnetizing

force dH produced by dl at a point P is dH ————————————————————————– r— …-(1)

The direction of this magnetizing force can be found by right-hand screw rule and it is directed into the place of the paper as shown by a cross at P.

Now,-                           xjIQ — dl.sxxiG

Also,                         ;c.sin# = r

or,                            jt = r/sin<?-                               ….(2)

where r is the perpendicular distance of P from the conductor AB.

-* 1.. 1 JCjdO IA9

dN ~——- — = —

4xx 4xx                                                             ■

2.6.                                                                                                                                                                                                                                                                                                                                                      An iron ring of mean length 50 cm has an air-gap of 1 mm and a winding of 200 turns. The relative permeability of iron is 300 when 1 amp currant flows through the coil. Determine the flux density.                                                                     [WBUT 2009(DECEMBER)]

Answer;                                                       .                                                                                         .

Given data                         :                                                .

Mean Length of iron ring                                      ‘

I                                                                                                                                                                                                                                                    – 50cm – 0.5m                                                                          .

I           ~lmm = lxl0“3m / =1 Amp.

Number of Turns N s= 200                                                                                             .

Relative permeability of iron //r =300

For finding the flux density, the ampere Turns for iron and air gap are to be determined> Leakage and fringing effects are neglected.

At iron *= H x / .

Where H is the magnetic field intensity

H = JL. =—————- — —— AT/m = 2652.585 AT/m

4;rxl0 x300                         –

where B is the flux density.

Hence AT for iron = HI AT = 2652.585×0.5 AT = #26.29 AT

AT for Air gap=                         AT ~ ^xlG’1x^~3 AT = 79577S A>r

Total ampere turns = B (1326,29 + 795.77) AT. = 2122.065 AT.                                                                         ■

Total ampere turns can be expressed as Nx l = 200×1 AT 200×1 = 2122.065

i . “i. ” ^5

Hence flux density B —————Wb/m2 = 0.094Wb/m2

J 2122.06 i

2.7.                                                                                                                                                                                                                                                                                                                                                    State Gauss’s law.                                                                     [WBUT 2009(DECEMBER>]

Answer: _______ _____

Gauss’s Law for B-fields

For E-fields Gauss’s law states in integral form that i EdS =

‘The flux of E over any closed surface is equal to the algebraic sum of the charges enclosed by the surface divided by e0 .*

In differential form Gauss’s law for E – fields has the form V-E = pfe0 where p is the free charge density. –

However for the B-field case there is no equivalent of the isolated charge, only magnetic dipoles which can be thought of as consisting of two equal but opposite magnetic monopoles. Hence the summation on the right hand side of Equation will consist of an equal number of positive and negative terms resulting in a total summation of zero.

Hence for B-fields, Gauss’s law has the form

BdS~0 and

3.1.                                                                                                                                                                                                                                                                                                                                                    a) i)Explain Faraday’s Law of electromagnetic induction.                                                                     [WBUT 2001]

OH                                                                            .                                                                            ■ 1 ,State and explain Faraday’s law of electromagnetic induction. [WBUT2004,2006]

ii)  State the difference between statically and dynamically induced e.m.f.s.

[WBUT 2001]


What is the difference between statically induced e.m.f. and dynamically induced

e.m.f.?                                                                                                  [WBUT 2004]

Answer:                  .                               . .

Faraday, in 1831, showed that, whenever the number of lines of magnetic flux Unking with an electric circuit is changed, an electromotive force is induced in the circuit, the magnitude of which is proportional to the rate of change of flux. Thus, if e is the e.m.f.

induced by a rate of change of flux of , e <x

dt dt

The e.m.f. is also proportional to the number of turns of wire, N, in the circuit in which the e.m.f. is induced.         .


Thus e N————– .


^                        r

If e is expressed in volts and 0> in webers, then e = -N—.


The negative sign is used to establish the fact that the e.m.f. induced always opposes the causet as stated in Lenz’s law,                                                                           ■

The unit of magnetic flux is defined by this relationship. The weber is the magnetic flux which, linking a circuit of 1 turn, produces in it an e.m.f. of 1 volt as the flux is reduced to zero at a uniform rate, in one second.

The phenomenon of electromagnetic induction plays a crucial role in the three most useful of all electrical devices: the electric generator, the electric motor, and the transformer,

4                                                                                                                                                                             – _

Self-Inductance with Reference to Faraday’s Law

Definition:                                       ■ ‘                                .                                                    .

Self Inductance: It is the property of an electric circuit or component that causes an e.m.f. to be generated in it as a result of a change in the current flowing through the circuit.      ■

The property of self-inductance is a particular form of electromagnetic induction. Self­inductance is defined as the induction of a voltage in a current-carrying wire when the current in the wire itself is changing. In the case of self-inductance, the magnetic field created by a changing current in the circuit itself induces a voltage in the same circuit. Therefore, the voltage is self-induced.

The term inductor is used to describe a circuit element possessing the property of inductance and a coil of wire is a very common inductor. In circuit diagrams, a coil or wire is usually used to indicate an inductive component.

The number of turns in the coil has an effect on the amount of voltage that is induced into the circuit Increasing the number of turns or the rate of change of magnetic flux increases the amount of induced voltage. Therefore, Faraday’s Law must be modified

for a coil or wire and becomes the following, y ~ jsf

’ * dt

Where:                  1 ‘                                   ‘

Vl = the induced voltage in volts N = the number of turns in the coil .

d<t>/dt = the rate of change in magnetic flux in webers per second,

The equation simply states that the amount of induced voltage (V^) is proportional to the number of turns in the coil and the rate of change of the magnetic flux (d0/dt).ln other words, when the frequency of the flux is increased or the number of turns in the coil is increased, the amount of induced voltage will also increase.                                                                         .

In a circuit, it is much easier to measure current than it is to measure magnetic flux so the following equation can be used to determine the induced voltage if the inductance and frequency of the current are known. This equation can also be reorganized to allow the inductance to be calculated when the amount of inducted voltage could be determined

. di ■ and the current frequency is known.. Vl = L

Where                         .

VL — the induced voltage in volts                                                                          .

L = the value of inductance in henries                                                                          .

di/dt = the rate of change in current in amperes per second. .

2nd Part:

») It follows from the law of electromagnetic induction that the instantaneous e.m.f.

induced in a coil                    e = ~N —                      –

dt ‘ ’

d<j> ‘ . . where —— is the time rate of change of flux and N is the number of turns of the coil dt .

If the exciting current is alternating in nature with the resultant flux being alternating and the machine is a static one like that of a transformer, the induced e.m.f. is termed as statically induced e.m.f. On the other hand, if the exciting current is direct current (e.g. field current of d.c. generators or alternators) and the field flux changes due to rotation of conductors in the magnetic field, the induced e.m.f. is referred to as dynamically induced e.m.f.                                                                                                       ‘                                                                                                      . ,

The expression for induced e.m.f. is always proceeded by negative sign because it follows from Lenz’s law that the direction of induced e.m.f. is such as to oppose the cause, i.e. the change in the inter linking flux.

For dynamically induced e.m.f., the direction of the e.m.f. is given quite simply by the right-hand rule. By this rule, the corresponding directions of motion, flux, and induced e.m.f. are given by the thumb, forefinger and middle finger respectively of the right hand, being so placed to be mutually perpendicular. The magnitude of the induced e.m.f. in a conductor which moves in a plane perpendicular to the flux is expressed as, e — flux per second = B x area swept by the conductor e = B/v, where B is the flux density in ‘Tesla’, I the length of the conductor in ‘m\ v the velocity of the conductor perpendicular to the field in m per sec, V is expressed in volts.

If a conductor cuts through magnetic flux while moving in a direction making an angle 6 with that of the lines of force, then the component of its velocity in a direction perpendicular to the field is v sin9, then the induced e.m.f. is                                                                                                      ‘

e~Blvsmd                                                                                           .

Energy in an electric circuit                    ,                                                                .

If the conductor in which the e.m.f. is induced forms a part of a closed circuit, and if a current of * units flows in this circuit, then there will be a force opposing the notion of the conductor through the magnetic field, which force is given as Bil.

Thus, the work done in moving the conductor a distance V through the field,

= Bilx.                                                                                             •

If the conductor takes a time t sec to pass through distance ‘at’ when movinjg with velocity v, then

work done =cBil vt — ei t = energy given to the electric circuit .*. Energy given to the electric circuit = eit.

b)                                                                                                                                                                                                                                                                                                                                                Two coils are packed on the same core. Their combined inductance is 0.9H when current through them flows in the same direction. The combined inductance falls 0.2H when current through them flow in opposite direction. The self inductance of one coil is 0.3H. Calculate values of (i) self inductance of other coil, (Ii) the Mutual inductance and (iii) the coupling coefficient.                                                                 [WBUT 2001 ]


Let Li and L2 be the self inductance of coil 1 and coil 2 respectively.

According to the problem, the equivalent inductance of the circuit when current flows in the same direction

Leq=L,+L2 + 2M……………… (i)

The equivalent inductance of the circuit when current flows in different direction L^ ~ L, + L^— 2M ……………………………………. (ii)……………………………………. –

Adding equations (i) and (ii)                                                                          _

2(£1+L!)=a9+a3 =L1

or.Zj +4=055; L, =03H .*. Lj =025//

(i)            Self inductance of the other coil = 0.25H

(ii)          the mutual inductance be

/. 2Af = 0.9—0.55 =0.35// or, M = 0.175//

(iii)               Let k be the coefficient of coupling

0,175 ■ = 0.639 V0.25 X 0.3

The coefficient of coupling = 0.64                                 .                                                                                                .

3.2.  a) Find the field strength at a point due to a current carrying conductor of finite length.

b)                                                                                                                                                                                                                                                                                                                                                An iron ring of mean circumference 50 cm has an air gap of 0.1 cm and a winding of 300 turns. If the permeability of iron is 400 when a current of 1 ampere flows through the coil, find the flux density in the sir gap.                                                                 [WBUT 2001]


a)   Refer to Question No. 2.5. Short Answer Type Questions.

b)     Mean circumference of the ring                                     ‘

lr = 50cm=50 xl0″2m = 0.5m                                                              1

Length of the air gap . .

L- = 0.1cm = 0.001m

Number of turns of the winding

JV = 300 ^=400

Let B wb/m2 be the flux density in the air gap.

2 .

Let Am be the cross sectional area of the ring

™ m. m. f Flux — —


Total ampere turns (m.m.f)                                     v

= Flux x reluctance of the iron – path + Flux x reluctance of the leakage and fringing,                            ■

or, B — 3x32^ x | q-4 wh f m2


3.3.  a) Explain how magnetic flux density can be defined from Ampere’s law.


State and prove Ampere’s Circuital law.                                                                                                          [WBUT 2003]

Write short note on Ampere’s circuital Law.                                                                                                          [WBUT 2005 2007 20081


■                                                                                                                                                                                                       ■ p (a) The above application of the Biot-Savart law showed that the B-field at a distance r

from an infinitely long, straight wire is B — —                                                                                                          ■

2 Kr

Consider a circular path L around the wire. The path has a length 27tr B

so the previous equation can be written as B2;cr = Pol or field x path

length — Pol. This result can be shown to be a special case of a general result.                                                                          .                                                                                                         Fig.6

B.dL = Ju0^I                                                                                             .

‘The line integral of B around a closed loop L is equal to the algebraic sum of the currents which flow through the area bounded by L multiplied by p*,.’

This result is known as Ampere’s Circuital law.

.. ■ _ .. _

any point on the line through the center and perpendicular to the plane of a circular current loop.                                                                      [WBUT 2002

magnetizing force (dH) at a point P, on the axis of the circle, due to the current in an

Fig: 1 above shows a circular conductor carrying j units of current. Consider the

Mi element dl of the conductor. Then                              dH

. . . . 4jtR2

in a direction perpendicular to the line joining the element to P.

This force may be split up into two components, one on direction OP produced, namely

dH sin# and one perpendicular to OP, namely dH cosfl Considering all such elements of

the circular conductor it is seen the components perpendicular to OP neutralize one

another, leaving, as the force at P, only the sum of all components in’ a direction OP

produced. Thus the total magnetizing force at P due to the current is in direction OP and is                                          .                                               •




At the center O of the circular conductor R = r,

3.4.  a) Derive an expression for the energy stored in a magnetic field. [WBUT 20021


Derive an expression for energy stored in a magnetic field.


Deduce an expression of energy stored of in a magnetic circuit. ,[WBUT 2O09(DEDCMBER)]OR

Derive the expression for energy stored in electric field.                                                                          [WBUT 2010{JUNE)]


a)     Energy Stored In the Magnetic Field of an Inductor


Fig: (i)                     ■

The above inductor having inductance L is connected to the battery through Key K. In this case inducted e m f is given by: .

, dt . e=~L* -<*> . . >

To drive the current through the inductor against the induced emf ‘e\ the external voltage is applied from the battery which has emf E and in this case E = – e.

Hence equation (1) becomes E — L.~—


Let an infinitesimal charge dq be driven through the inductor. So the work done by the external voltage to do this work is given by:

dw — E dq

a i di , T j { dqN

or, dw = L—.dq — L.di

■             dt


dq _


dw = Li di

Total work done to maintain the maximum value of current (lo ) through the inductor is given by

V” h


— L = L.



™ —JO

_ 2 _

The work done in increasing the current flowing through the inductor is stored as the

potential energy (U) in its magnetic field. U =— LI2

b)                                                                                                                                                                                                                                                                                                                                                                Two coils, with a coefficient of coupling of 0.5 between them, are connected in series so as to magnetise (a) in the same direction and (b) in the opposite direction. The total inductance for connection (i) is 2.0H and for connection (ii) is 0.8H. Find the self-inductance of the two coils and mutual inductance between them.                                                                                            [WBUT 2002]


Let Li , L2 and M be the values of self-inductance and mutual inductance respectively.

(i)               For the first case when they are connected in such a manner to magnetise in the same

direction, Leq = L, 4- + 2M — 2.0…………… (i)

(ii)               When they magnetise in different direction

Leq – Li + L2 — 2M -0.8…………… (ii)

Adding equations (i) and (ii) we get                         .

L, + L2 = = 1.4H ……(iii)


2M = 2 -1.4 = 0.6H

Therefore, mutual inductance M = 0.3H                                  ■

Now, M =

where k is the coefficient of coupling jfc = 0.5

k 05 LjZ^ — 0.36

4/^ =. 1.44


(Ll + L!)1^(L,-Lif+4L,L2 (L, — Lj)2 == (Lj + £2)* —41^1*2 — (1.4)2 —1.44 =-1.96—1.44 =052

, L, — L2 = -J052 = 0.72……………. (iv)

Adding equations (iii) and (iv) we get 2Lt=212H; L, = 1.06//

4 – (1.4 -1.06)// = 0.34//

Ans. The self-inductance of the two coils are L06H and 0.34H. The mutual inductance of the coil is 0.3H.

3.5.                                                                                                                                                                                                                                                                                                                                                      a) Draw an expression for lifting force of an electromagnet.                                                                                                   [WBUT 2003]

b)        A horso Shoe electromagnet is required to lift 200Kg-weight. Find the exciting Current required if’the electromagnet is wound with 500 turns. The magnetic length of the electromagnet is 60cm and is of permeability 500. The reluctance of the load can be neglected. The pole face has a cross section 20sq.                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                            Lifting force of an electromagnet                                   v .

Consider two poles of a horse-shoe magnet, as shown in the following fig. 1

Let the surface-area of each pole be equal to A sq-m.; and let F be the force of attraction between the pair of poles.                                                                                               –                                                                                                .

Let one of the poles be displaced by an incremental distance Sx against the force of attraction; while the exciting current be assumed constant The work done in the process, against force F = F6x Joules

The volume of the magnetic field between the poles is increased by A8x cu-m. So, the energy stored is increased by Energy density of the mag.                                                          ’

Field x A5x Joules = — (fi2 / ) xA 5x Joules                                                                                              

Lifting power of an electromagnet is numerically equal to —-

the flux density in Wb/m2, A is the surface area in m2, fl is the relative permeability of the magnetic medium, and /U0 is the permeability of free space.

b)         Let I Ampere be the required current.

Weight to be lifted = 200 Kg-weight Therefore, force required to lift the required weight = 200 X 9.8N = 1960N

Total finv generated – M M J(AmPen                                                                                                         ;                                                                                                         .

Reluctance of the iron path                –

, 500fxunu

where A is the cross-sectional area in m2                         ‘

Therefore, Flux density B = SOO^^xlQ-’xSOO

‘ 0.6 :

A — 20cm2 = 20x 10^m2




(500) x/2x(4^x!0-7) x(500)2 X 20×10

2         x (0.6)x x 4jtx 10 7 x 500

(500) x /2 x 20 x 10^ x 4^x10                                 2

or,±————————————————– = 9.8 X –

2                                                                                                                                                                                                                                          ■ 5

= 9.8 x 0.4

(0.6)2 9j8 x (0.6)2


or, I = 0.6 X10 2 xj“ x 0.4A = 67A

3.6.  a) Compare magnetic circuit with electric circuit

Compare electric and magnetic circuits with respect to their sim^rrties and dissimilarities.                                                                       [WBUT 2010(DECEMBER)]


Magnetic circuits

f~ Electrical circuits 1
* ………. \

‘ >

——— l|lj^—, . –
.an: < iMMF


J – ■ ” R



* –


‘ Magnetic circuits .

Electrical circuits
  1. Flux flows from N-pole to S-pole
  2. Flux flow is due to mmf.
  3. Resistance to flow of flux is offered by reluctance.

mmf Flux =


(i)                              mmf is expressed in ampere turns.

(ii)               Flux is expressed in webers.

(iii)                             Flux density is expressed in Wb/m2 …….

  1. Reluctance R = —— where fi is



  1. Permeance =—————- * ■

i reluctance ■ – •


6. Permeability —————– ——


Current flows from +ve to —ve Current flows due to e.m.f. Resistance to flow of current is offered by resistance R. .’ ■ emf Current ————-

. resistance

(i)             emf is expressed in volts

(ii)               Current is expressed in amperes.

(iii)                      Current density is expressed in (A/m2).

Resistance R = ——

a A

where <7 is the conductivity.

Conductance ———— *——


Conductivity =———————

■ resistivity .

b)       Calculate the mmf required to produce a flux of 0.01 Wb across an air-gap of 2 mm of length having an effective area of 200 cm2 of a wrought iron ring of mean iron path of 0.5 m and cross-sectional area of 125 cm2. Assume a leakage coefficient of 1.25. The magnetization curve of the wrought iron is given below:

.                 B (Wb / m ): 0.6 0.8 1.0 1.2 1.4 .

H (AT / m): 75 125 250 500 1000                ‘

,               {WBUT 2006}

Answer:                                                                        .

For iron path

”                                                                                                                                                                                                                                                                                                a .                                                                                   n                                                                                                                                                                                                                                                                                                                                                   ;                                                                                                                                                                                                                                                                                                  ’

Cross-sectional area A = 125 cm Mean Length L = 0.5 in

Considering leakage coefficient of 1.25,                                           .

Flux density                            B —   Wb/m2 = 1.0 Wb/tn2

125×10                                            ..                          .               .

for    B = 1.0’Wb/m2, H — 250 AT/m

. Ampere turns necessary                                                  •

= HI = 250×0.5 AT = 125 AT

For air gap, area                A = 200cm2 — 200×10″”* m2                                   •

+ * ii h

mean length= 2mm — 2xl0-3 m ‘ ‘                                                                                                         – –                                                                                                         .

1 25×0 01 ‘

Flux Density B =

Wb/m2 = 0.625 Wb/m2

200×10 1 ‘ b 0 625

Intensity of magnetisation H == —■— ———————————————————— :AT/m for

fiafir 4^rxl0 xl air =497359.2 AT/m ‘ mmf – HI = 397887.36x2xl0~3 AT = 994.7184 = 995 Hence, total mmf required = (125+995) A=1120 AT

3.7.                                                                                                                                          Define self and mutual inductance. What do you mean by co-efficient of coupling?                            [WBUT 2007, 200S(DECEMBER)&(JUN£), 2009(DECEMBER)JOR

Define self-inductance and mutual inductance.                                                                                              [WBUT 2009(DECEMBER)]OR

Two coils have self inductances Z, and and mutual inductance between them

is ML Derive a mathematical expression for co-efficient of coupling k for these coils.                                                                                             [WBUT 2011 (DECEMBER)]

Answer:                                                                                            .

Definition:                                                   .

Self Inductance: It is the property of an electric circuit or component that causes an e.m.f. to be generated in it as a result of a change in the current flowing through the circuit.

The property of self-inductance is a particular form of electromagnetic induction. Self­inductance is defined as the induction of a voltage in a current-carrying wire when the current in the wire itself is changing. In the case of self-inductance, the magnetic field created by a changing current in the circuit itself induces a voltage in the same circuit. Therefore, the voltage is self-induced. .

The term inductor is used to describe a circuit element possessing the property of inductance and a coil of wire is a very common inductor. In circuit diagrams, a coil or wire is usually used to indicate an inductive component.                                                                                                         –

The number of turns in the coil has an effect on the amount of voltage that is induced into the circuit. Increasing the number of turns or the rate of change of magnetic flux increases the amount of induced voltage. Therefore, Faraday’s Law must be modified

for a coil or wire and becomes the following. VL= N ——


VL = the induced voltage in volts                           –

N = the number of turns in the coil                                                                                                          .

d<|>/dt = the rate of change in magnetic flux in webers per second

The equation simply states that the amount of induced voltage (VjJ is proportional to the number of turns in the coil and the rate of change of the magnetic flux <d((>/dt).In other

words, when the frequency of the flux is increased or the number of turns in the coil is increased, the amount of induced voltage will also increase.

In a circuit, it is much easier to measure current than it is to measure magnetic flux so the following equation can be used to determine the induced voltage if the inductance and frequency of the current are known. This equation can also be reorganized to allow the inductance to be calculated when the amount of induced voltage could be determined and

. di ■ the current frequency is known. VL = L—~ .



VL = the induced voltage in volts                                                                          .

L = the value of inductance in henries                                   .

di/dt = the rate of change in current in amperes per second ‘

Mutual Inductance & Co-efficient of Coupling

The magnetic flux through a circuit can be related to the current in that circuit and the

currents in other nearby circuits, assuming that there are no nearby permanent magnets.

[Note: Magnetic flux is a measurement of quantity of magnetism, taking account of the

strength & the extent of the magnetic field.]

—– fr———- AAA

Consider the following two circuits:

The magnetic field produced by circuit 1 will link the wire in circuit 2 and induce emf in circuit 2 causing a current to flow. The induced current flow in circuit 2 will have its own magnetic field, which will interact with the magnetic field of circuit 1. At some point P on the magnetic field consists of a part due to ii and a part due to it. These fields are proportional to the currents producing them.

The coils in the circuits are labeled Li and Lj and this term represents the self-inductance of each of the coils. The values of L| and L2 depend on the geometrical arrangement of the circuit (i.e. number of turns in the coil) and the permeability of the material. The constant M is called the mutual inductance of the two circuits and it is dependent on the geometrical arrangement of both circuits. In particular, if the circuits are far apart, the magnetic flux through circuit 2 due to the current i1 will be small and the jautual inductance will be small. L2 and M are constants.                                                                                                        –

We can write the flux, <pB through circuit 2 as the sum of two parts,                                                                                                          .

0B2= + hW                           •

An equation similar to the one above can be written for the flux through circuit 1.

fist ~ ^tii +

Though it is certainly not obvious, it can be shown that the mutual inductance is the same for both circuits. Therefore, it can be written as follows:

Mutual Inductance:                                                                    ‘

Let there be two coils A and B as shown in the figure

r*                                                                                                                       jjr m



‘ – o———————                           _

Let, La Jind Lb be the self — inductance of coil A and B respectively.

In the given case resultant flux is subtractive when current flows in both coils.


• The voltage induced in coil A \eA\- LA d\ fdt~M d^/dt

Where M is the mutual inductance voltage induced in coil B.

The voltage induced in coil B \eB\~ LB di^jdt ~M d\fdt

If the dots are placed in such a way that the resultant flux is additive when current flows, the induced voltages will be as follows:                                                                           ’

|®a| “ La dij/dt + Mdi2/dt eB| = LB di2/dt + Mdij/dt

If both the coils are wound on the same core such that they magnetise in the same way when current flows through shown (i.e. the resultant flux is additive), the induced emf . . e = difdt + M dijdt + L, dijdt + M dijdt

Where L, & L2 are the self – inductances and M is the mutual inductance.

Equivalent inductance = Li+La + 2M.

When the orientation of the coils is such that the resultant flux is subtractive, e-L^ di/dt + difdt – 2M dijdt                                        ’

Equivalent inductance = Lt + L2 – 2M

Co-eflkJent of coupling:                 ‘                                •

The mutual inductance gives an indication of the amount of closeness or proximity of the coupling betwisiil the two coils.

Co-efficient of coupling provides a quantitative measure of the magnetic coupling of two coils whose self-inductances are Lj and Lq respectively.                                                                   ,

  • Mathematically, K ~

VAA                                        *

. n ■ ■ .

Where M is the mutual inductance. Maximum value of K can be unity when the entire magnetic flux set up by coil L3 is linked with all the turns of the coil L2.

3.8. a) Derive an expression for the lifting power of an electromagnet


Answer:                                    .

Lifting power of an electromagnet

Consider two poles arranged directly opposite each other as shown
in the figure a. Let each, have an area ‘A’ square metres and let F
Newtons be thevforce of alteration between them. Let one pole be

moved by an incremental distance Sx against the force F- then the
work done is obviously F?Sx Joules.

The volume of the magnetic field between the poles is increased by
ASx cu-m.                                                                               ‘ *

■                                                                                                                                                                                                                                                                              L

Hence, the energy stored is increased by
Energy density of the magnetic field x ASx Joules


Fig: a

b) A horseshoe electromagnet is required to lift a load of 200 kg. The iron length of the horseshoe magnet and the load is 60 cm and of relative permeability 600. Find the current drawn by the electromagnet, if ft is wound with 1000 turns. The contact gap of the load and magnet is 0.001 cm at each pole. The cross-section of the pole is 20 sq. cm. Assume no leakage of Mux.                                                                                                    [WBUT 2008(DECEMBER)]

Answer:                                                                                                                                  ’                              . ■

The force exerted by each pole of the electromagnet to lift the load

f = ™Xg x9.8IN = 981N                                                                                                         •

■ 2 2 ■

The cross-sectional area of the pole

A = 20×10-4 Sq.ra = 0.002 Sq.m                                        –

, where — 1 for air

Flux density B = yJlFJ/jA = J2×4**VL-

Intensity of magnetization

B                    1.11

= 1472.578 AT/m

4jtx10 7 x600 Length of iron path I = 60cm = 0.6 m

Required mmf = 1472.578 x 0.6 = 883.547 AT

No. of turns — 1000                                                \                                                                                                       •                                                                                                       – –

, m.m.f 883.547 n oooc A current ——————- ——————- — 0.8835 A no.of turns 1000


3.9.          A horse-*hoe electromagnet is required to lift a load of 200 kg. The electromagnet is wound with 500 turns and the length of the magnetic path is 50 cm and the cross- section of each arm is 25 sq. cm. Find the current in the coil.

State the assumptions. [jJ.r =400]                                                                      [WBUT 2008(JUNE)]

Answer:                                               .

The assumptions taken are

i)          There is no leakage flux and the effect of fringing is negligible.

ii)           The flux density is same throughout.                               –

iii)                 The magnetic characteristics of the material are the same throughout the entire

volume of the core.                                                 –

The force exerted by each pole of the electromagnet.

– 200 . K = —xg = 100×9.81 Af — 981N ‘ 2

Cross-sectional area of each pole

A = 25 sq.cm = 25×1 O^m2 = 0.0025m2

Intensity of magnetization, H — —*— = ————- T~^j—TTTTT = 1975.67 AT/rtl

Bm             0.993

w ____  _____  ____

/J0/Ir 4^x10~7x400

Length of magnetic path = 50 cm = 0.5 m Required mmf — 1975.67 x 0.5 = 987.83 AT

KT * required m.m.f 987.83 ,t Q_,, A Necessary current = ———–     —————————————— —■ =- = 1.9756 A

No. of turns 500

■ * ii ’ i

3.10.                                                                                                                                                                                                                                                                                                                                             a) A cast steel ring has a cross-section area of 7.5 sq cm & a mean length with a circumference of 75 cm. The ring is uniformly wound with 900 turns. Find out the current required to produce a flux density of 1 Wb/m2 in fhe ring if the relative permeability of cast steel is 1500. If a saw cut of 1.5 mm is made in the ring, find out the current required to give the same flux density in the ring. ‘

b)                                                                                                                                                                                                                                                                                                                                                                     Two colls having 3000 A. 2000 turns are wound on a magnetic ring. 60% of flux produced in first field coll links with second coil. A current of 3A produces flux of 0.5 mWb in the first coil & 0.3 mWb in the second coil. Determine the mutual inductance & the coefficient of coupling.                                                                          ‘ [WBUT 2010(JUNE)]


a)     A cast steel ring has a cross-sectional area of 7.5cm2 = 7.5xl0~4m


Length of iron path = 75 cm = 0.75 m

Flux density = 1 Wb / m2                                               ‘

fir =1500                                                                       ,

= 4/TXlO”?

#                                                                                                                                              ..

Intensity of magnetization = V = 1/ „      AT,

/MjA /4^rxl0 xl500 /m

105 10000 j

60# 6/r /m

10 000 ■ ’ Necessary Ampere turns =—1————– x0.75 AT

Let 1 be the current flowing Number of turns = 900

900/-B222x0.75=™°                                                              ’

6sr                           –

75                                                                                                                                                 ‘

/ = = 0.44 A


If a saw cut of 1.5 mm = 1.5xl03 mis made in the ring then ampere turns for air gap

AF*S                 ~ 4#xl0-7 xL5xW = 4jl AT = U93.7 AT.


Total ampere turns =


= (397.89 +1193.7) AT = 1591.59 AT.

1591.59 — 900×7 / = 1.77 A

Current required = 1.77A *>)

Nx =3000 N2 =2000 ft = 0.5 mWb

^=0.3 mWb /,=/2=3A                       ’

<p2 -0.6$ – . . ” . .


=0.2 H

ht _ JVi & _ 3000×0.5xl0~3 , „

a‘”7                      3 :=aSH

, _N2& 2000×0.3×103 „„„ t, —————–

= .6324


3.11.  a) State and explain Biot-Savart law.


Refer to Question NoL 2.5. Short Answer Type Questions;

b)          A ring having a mean diameter of 21 cm and a cross-section of 10 cm2 is made of two semicircular sections of cast iron and cast steel respectively with each joint having reluctance equal to air gap of 0.2 mm as shown in figure. Determine the ampere turns required to produce a flux of 0.8 mWb. The relative permeabilities of cast iron and cast steel are 166 and 800 respectively. Neglect fringing and leakage effects.


From the given data:

Diameter of the ring                                                                        .

D – 0.21m

~ Cross-sectional area A — 10×10 4 m2 = 10 3m2 Lagging air gap

lag =0.2mmx2 = 0.2x2xKT3m =0,4xl0″3m


4#xl0″7xl66 2 0.8×0.21xl07

AT = 1265 AT 8×166

The ampere turns for the cast steel path

= Hx£ =                                x —–1

Aitx 10 x800 2 8 100

= -x 8 8

Ampere turns for air gap


= —x2x0.2xl0~3 AT


= ———=-x0.4xl0~3 AT 4^x10 _ 0.8 x 0.4×104


 = —X—xlOO AT                                        .                                                                  ,

—                                         AT = 254.65 AT – ‘


Total ampere turns necessary = 1265 + 262.5 + 254.65 = 1782 AT

3.12.                                                                                                                                                                                                                                                                                                                               A flux of 0.0006 Wb is required in the air-gap of an iron ring of cross-section 5.0 cm2a and mean length 2.7 m with an air-gap of 4.5 mm. Determine the ampere turns required. Six H values and corresponding B values are noted from the magnetization curve of iron and given below.                                                                  [WBUT 2011 (DECEMBER)]

H (AT/m) 200 400 500 600 800 1000
B (Wb/m2) 0.4 0.8 1.0 1 1.09 _ 1? , 1.19


Given data:

Flux = 0.0006 Wb.

Cross sectional area A — 5.0cm2 ~ 5xl0~*m . ■ …. 0.0006,

= 1.2 Wb/m


From the given data

H(ATfm) for B = 1.2Wb/m2 =1000AT/m.

Hence ampere turns required for the iron part producing the given flux density

—HI s= 1000×2 7A7T =2100AT

Extra ampere turns necessary to produce the same flux in the air gap is calculated below.

B = 1.2 Wb/m2; H = B/fi0

AT / m=—— 1-^—AT/m=954930AT——————– ‘

4ttx10                                                                   ‘

Extra ampere turns required

= 954930X4.5xlO’3 AT = 4297.185A7\

Hence the total no. of ampere turns necessary

= (2700 + 4297.185) AT =6997.185AT

It is to be noted that the problem has been solved neglecting leakage & fringing.

3.13.                                                                                                                                                                                                                                                                                                                                              Write short note on Eddy current losses. •                                                                                                 [WBUT 2003, 2005]


Eddy Current Losses

Eddy current losses occur vyhenever the core material is electrically conductive. Most ferromagnetic materials Contain iron: a metal that has fairly low resistivity .The solid iron core is subjected to varying flux fields. So ‘eddy emfs’ are induced in the core and the solid core serves as a conducting path.

In any resistive circuit the power is proportional to the applied voltage. The induced voltage is itself proportional to fxB and so the eddy losses are proportional to fB also. The flux is also related to the size of the loop. The same principle applies to motors and generators too. Usage of a solid iron core results in large circulating currents. So, instead, the core is made up of a stack of thin (-0.5 millimetre) sheets (cross section C). The I ines of magnetic flux can still run around the core within the plane of the laminations. The situation for the eddy currents is different. The surface of each sheet carries an insulating oxide layer formed during heat treatment This prevents current from circulating from one lamination across to its neighbours.

Power loss (the reduction of which is our aim) is proportional to the square of induced voltage. Induced voltage is proportional to the rate of change of flux, and cach of our laminations carries one quarter of the flux. So, if the voltage in each of the four laminations is one quarter of what it was in the solid core then the power dissipated in each lamination is one sixteenth the previous value.                                                                                                        (

The eddy current losses of the core are expressed as Pe — kt f1

where ke is the proportionality constant whose value depends on the volume and

resistivity of the core material, thickness of the laminations and the units employed.

Bm is the maximum flux density of the corc and / is the frequency of the alternating


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