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 Ntums the selfinductance will be proportional to –
a) N b) — c) N^{2} 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) crosssection 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 ft_{r}A
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) crosssectional area and length
c) resistivity d) crosssectional area
Answer: (d) .
L
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)
iiTorm
2Wb/m‘^{t} (Tesla). If the conductor moves with a velocity of 5dm see^{1}, 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]
Answer:
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 aircored toroidal coil has 450 turns and a mean diameter of 300 mm and crosssectional area of 300 mm^{2}. Determine the selfinductance of the coil and the average voltage induced in it when a current of 2A is reversed in 40 ms.
{WBUT 2005]
Answer:
The number of turns of the ceil N = 450
Effective length i = &d = ttx300mm = 942.478mm = 0.942m
Cross sectional area A = 500 mm^{2} = 300 xlO’^{6} m^{2} (i = Permeability of air = permeability of free space
= M_{r}M_{0}= lx4*rxl(T^{7} _ # – Therefore m selfinductance of the coil ■ ■
, N^{2}/iA 450x450x4;rx300xl(r^{6}xICr^{7}
^{L}~——————————————————— ———— = 8l.041{iH
£ 0.942 –
The average voltage induced in an inductive coil
_{T}di 81.04×10^x2 ■ ‘ ’ ‘ ‘
^{v}t^{s=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]
Answer:
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 shortcirculated 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)]
Answer:
Hysteresis Loss
Flux density <B) 
TThe Hysteresis Loop 
P* 
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 P_{2}. 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 P_{3}. 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 (P_{4}). 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
loop.
The expression of Hysteresis loop can be written as W_{h} = K_{U}V fB^{16}W
Where K_{k} 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, B_{m} is the maximum value of flux density in the core in Wb/m^{2}
2.5. State BiotSavart law applicable to electromagnetism. [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 Bfield 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 — — –
4nr^{2}
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 ‘BiotSavart 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} Hiti^{1} (Henry/meter). _{:}
The unit of B is the tesla (T)” • ^ . .*? ,
By use of a suitable integration the BiotSavart law can be used to calculate the Bfield resulting from a circuit L that can be decomposed into an infinite number of connected
current elements B = cf ~.
Applications BioSavart Law . .
BioSavart 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 BioSavart 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 righthand 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 airgap 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“^{3}m / =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’^{1}^{x}^~^{3}^{ AT = 795}–^{77S 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/m^{2} = 0.094Wb/m^{2}
^{J} 2122.06 i
2.7. State Gauss’s law. [WBUT 2009(DECEMBER>]
Answer: _______ _____
Gauss’s Law for Bfields
For Efields 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 e_{0} .*
In differential form Gauss’s law for E – fields has the form VE = pfe_{0} where p is the free charge density. –
However for the Bfield 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 Bfields, 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]
OR
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. .
d<S>
Thus e N————– .
dt
^ r
If e is expressed in volts and 0> in webers, then e = N—.
dt
The negative sign is used to establish the fact that the e.m.f. induced always opposes the cause_{t} 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 – _
SelfInductance 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 selfinductance is a particular form of electromagnetic induction. Selfinductance is defined as the induction of a voltage in a currentcarrying wire when the current in the wire itself is changing. In the case of selfinductance, the magnetic field created by a changing current in the circuit itself induces a voltage in the same circuit. Therefore, the voltage is selfinduced.
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.. V_{l} = L
Where .
V_{L} — the induced voltage in volts .
L = the value of inductance in henries .
di/dt = the rate of change in current in amperes per second. .
2^{nd} 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 righthand 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 =^{c}Bil 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 ]
Answer:
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
L_{eq}=L,+L_{2} + 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]
Answer:
a) Refer to Question No. 2.5. Short Answer Type Questions.
b) Mean circumference of the ring ‘
l_{r} = 50cm=50 xl0″^{2}m = 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/m^{2} be the flux density in the air gap.
2 .
Let Am be the cross sectional area of the ring
™ m. m. f Flux — —
Reluctance
Total ampere turns (m.m.f) ^{v}
= Flux x reluctance of the iron – path + Flux x reluctance of the leakage and fringing, ■
or, B — ^{3}9°^{x32}^ _{x}  _{q}4 _{wh f m}2
18
3.3. a) Explain how magnetic flux density can be defined from Ampere’s law.
OR
State and prove Ampere’s Circuital law. [WBUT 2003]
Write short note on Ampere’s circuital Law. [WBUT 2005 2007 20081
Answer:
■ ■ p (a) The above application of the BiotSavart law showed that the Bfield 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 _{= J}u_{0}^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 . •
4xR^{2}
ix2xr
4jiR^{1}
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
OR
Derive an expression for energy stored in a magnetic field.
OR
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)]
Answer:
a) Energy Stored In the Magnetic Field of an Inductor
dl
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.~—
dt
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 { dq^{N}
or, dw = L—.dq — L.di
■ dt
j
dq _
dt
dw = Li di
Total work done to maintain the maximum value of current (l_{o} ) through the inductor is given by
V”  h 
I^{2} 

— L  = L. 
0 

2 
™ —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 =— LI^{2}
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 selfinductance of the two coils and mutual inductance between them. [WBUT 2002]
Answer:
Let Li , L_{2} and M be the values of selfinductance and mutual inductance respectively.
(i) For the first case when they are connected in such a manner to magnetise in the same
direction, L_{eq} = L, 4 + 2M — 2.0…………… (i)
(ii) When they magnetise in different direction
Leq – Li + L_{2} — 2M 0.8…………… (ii)
Adding equations (i) and (ii) we get .
L, + L_{2} = = 1.4H ……(iii)
2
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
Now,
(L_{l} + L_{!})^{1}^(L,L_{i}f+4L,L_{2} (L, — Lj)^{2} == (Lj + £2)* —41^1*2 — (1.4)^{2} —1.44 =1.96—1.44 =052
, L, — L_{2} = 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 selfinductance 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 200Kgweight. 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 horseshoe magnet, as shown in the following fig. 1
Let the surfacearea of each pole be equal to A sqm.; 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 cum. So, the energy stored is increased by Energy density of the mag. ’
Field x A5x Joules = — (fi^{2} / ) xA 5x Joules
Lifting power of an electromagnet is numerically equal to —
the flux density in Wb/m^{2}, A is the surface area in m^{2}, fl is the relative permeability of the magnetic medium, and /U_{0} is the permeability of free space.
b) Let I Ampere be the required current.
Weight to be lifted = 200 Kgweight Therefore, force required to lift the required weight = 200 X 9.8N = 1960N
Total finv generated – ^{M M}^{ J(}^{Am}P^{en} ; .
Reluctance of the iron path –
, 500fxu_{n}u
where A is the crosssectional area in m^{2} ‘
Therefore, Flux density B = SOO^^xlQ’xSOO
‘ 0.6 :
A — 20cm^{2} = 20x 10^m^{2}
B?A
2Wo
2
(500) x/^{2}x(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}
jtxIO

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)]
Answer:
Magnetic circuits 
f~ Electrical circuits 1  
* ………. \ 
‘ > 
——— llj^—,  . – 
.an: < iMMF
* ^{J} – ■ ” R 
I^{1}!^{1} ■ RESISTANCE * – 
/ 
‘ Magnetic circuits . 
Electrical circuits 
mmf Flux = reluctance (i) mmf is expressed in ampere turns. (ii) Flux is expressed in webers. (iii) Flux density is expressed in Wb/m_{2} …….
jiA permeability.
i reluctance ■ – • 1 6. Permeability —————– —— reluctivity 
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/m^{2}). Resistance R = —— a A where <7 is the conductivity. Conductance ———— *—— resistance Conductivity =——————— ■ resistivity . 
b) Calculate the mmf required to produce a flux of 0.01 Wb across an airgap of 2 mm of length having an effective area of 200 cm^{2} of a wrought iron ring of mean iron path of 0.5 m and crosssectional area of 125 cm^{2}. 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 ^{;} ’
Crosssectional area A = 125 cm Mean Length L = 0.5 in
Considering leakage coefficient of 1.25, .
Flux density B — Wb/m^{2} = 1.0 Wb/tn^{2}
125×10 .. . .
for B = 1.0’Wb/m^{2}, H — 250 AT/m
. Ampere turns necessary •
= HI = 250×0.5 AT = 125 AT
For air gap, area A = 200cm^{2} — 200×10″”* m^{2} •
+ * ii h
mean length= 2mm — 2xl0^{3} m ‘ ‘ – – .
1 25×0 01 ‘
Flux Density B =
Wb/m^{2} = 0.625 Wb/m^{2}
200×10 ^{1} ‘ b 0 625
Intensity of magnetisation H == —■— ———————————————————— ^{:}AT/m for
fi_{a}fi_{r} 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 coefficient of coupling? [WBUT 2007, 200S(DECEMBER)&(JUN£), 2009(DECEMBER)JOR
Define selfinductance 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 coefficient 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 selfinductance is a particular form of electromagnetic induction. Selfinductance is defined as the induction of a voltage in a currentcarrying wire when the current in the wire itself is changing. In the case of selfinductance, the magnetic field created by a changing current in the circuit itself induces a voltage in the same circuit. Therefore, the voltage is selfinduced. .
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. V_{L}= N ——
Where:
V_{L} = 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. V_{L} = L—~ .
dt
Where
V_{L} = 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 & Coefficient 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 selfinductance 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 i_{1} will be small and the jautual inductance will be small. L2 and M are constants. –
We can write the flux, <p_{B} 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}
ll
A
‘ – 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 \e_{A}\ L_{A} d\ fdt~M d^/dt
Where M is the mutual inductance voltage induced in coil B.
The voltage induced in coil B \e_{B}\~ L_{B} 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 “ L_{a} dij/dt + Mdi_{2}/dt e_{B} = L_{B} di_{2}/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, eL^ di/dt + difdt – 2M dijdt ’
Equivalent inductance = L_{t} + L2 – 2M
CoeflkJent of coupling: ‘ •
The mutual inductance gives an indication of the amount of closeness or proximity of the coupling betwisiil the two coils.
Coefficient of coupling provides a quantitative measure of the magnetic coupling of two coils whose selfinductances 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 L_{3} is linked with all the turns of the coil L_{2}.
3.8. a) Derive an expression for the lifting power of an electromagnet
[WBUT 2008(DECEMBER)] Answer: . Lifting power of an electromagnet Consider two poles arranged directly opposite each other as shown moved by an incremental distance Sx against the force F then the The volume of the magnetic field between the poles is increased by ■ L Hence, the energy stored is increased by 





, where — 1 for air
Flux density B = yJlFJ/jA = J^{2×4}**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
P
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
Crosssectional area of each pole
A = 25 sq.cm = 25×1 O^m^{2} = 0.0025m^{2}
Intensity of magnetization, H — —*— = ———— T~^j—TTTTT ^{=} 1975.67 AT/rtl 
B_{m} 0.993
w ____ _____ ____
/J_{0}/I_{r} 4^x10~^{7}x400
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 crosssection 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/m^{2} 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)]
Answer:
a) A cast steel ring has a crosssectional area of 7.5cm^{2} = 7.5xl0~^{4}m
I
Length of iron path = 75 cm = 0.75 m
Flux density = 1 Wb / m^{2} ‘
fi_{r} =1500 ,
= 4/TXlO”?
# ..
Intensity of magnetization = V = 1/ „ AT,
/MjA /4^rxl0 xl500 /m
10^{5} 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/B222_{x0}.75=™° ’
6sr –
75 ‘
/ = = 0.44 A
54;r
If a saw cut of 1.5 mm = 1.5xl0^{3} mis made in the ring then ampere turns for air gap
^{AF}*_{S} ~ 4#_{xl0}7 ^{xL5xW} = _{4jl} ^{AT} = U93.7 AT.
7500
Total ampere turns =
6tt
= (397.89 +1193.7) AT = 1591.59 AT.
1591.59 — 900×7 / = 1.77 A
Current required = 1.77A *>)
N_{x} =3000 N_{2} =2000 ft = 0.5 mWb
^=0.3 mWb /,=/_{2}=3A ’
<p_{2} 0.6$ – . . ” . .
2000×0.3xl0~^{3}
=0.2 H
ht _ JVi & _ 3000×0.5xl0~^{3} , „
^{a}‘”7 3 ^{:}—^{=aSH}
, _N_{2}& 2000×0.3×10^{3} „„„ t, —————–
= .6324
5×02
3.11. a) State and explain BiotSavart law.
Answer:
Refer to Question NoL 2.5. Short Answer Type Questions;
b) A ring having a mean diameter of 21 cm and a crosssection of 10 cm^{2} 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.
Answer:
From the given data:
Diameter of the ring .
D – 0.21m
~ Crosssectional area A — 10×10 ^{4} m^{2} = 10 ^{3}m^{2} Lagging air gap
l_{ag} =0.2mmx2 = 0.2x2xKT^{3}m =0,4xl0″^{3}m
_{x£}x02l
4#xl0″^{7}xl66 2 0.8×0.21xl0^{7}
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
B
= —x2x0.2xl0~^{3} AT
Mo
= ———=x0.4xl0~^{3} AT 4^x10 _ 0.8 x 0.4×10^{4}
4tt
= —X—xlOO AT . ,
— AT = 254.65 AT – ‘
JT
Total ampere turns necessary = 1265 + 262.5 + 254.65 = 1782 AT
3.12. A flux of 0.0006 Wb is required in the airgap of an iron ring of crosssection 5.0 cm^{2}a and mean length 2.7 m with an airgap 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/m^{2})  0.4  0.8  1.0 ^{1}  1.09  _ _{1?} ,  1.19 
Answer:
Given data:
Flux = 0.0006 Wb.
Cross sectional area A — 5.0cm^{2} ~ 5xl0~*m . ■ …. 0.0006,
= 1.2 Wb/m
5×10′
From the given data
H(ATfm) for B = 1.2Wb/m^{2} =1000AT/m.
Hence ampere turns required for the iron part producing the given flux density
—HI s= 1000×2 7A7^{T} =2100AT
Extra ampere turns necessary to produce the same flux in the air gap is calculated below.
B = 1.2 Wb/m^{2}; H = B/fi_{0}
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]
Answer:
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 P_{e} — k_{t} f^{1}
where k_{e} is the proportionality constant whose value depends on the volume and
resistivity of the core material, thickness of the laminations and the units employed.
B_{m} is the maximum flux density of the corc and / is the frequency of the alternating
supply.