EXERCISE
MULTIPLE CHOICE QUESTIONS
Choose the best possible answer:
1. For inducing e.m.f in a coll the basic requirement is that
a. Flux should link the coil
b. Change in flux should link the coil
c. Coil should form a closed loop
d. both (B) and (C) are true
2.When a single turn coil rotates in a uniform magnetic field, at uniform speed the induced e.m.f will be
a. Alternating
b. Steady
c. Pulsating
d. Zero
3.Pinciple of dynamically induced e.m.f is used in a.
a. Choke c. b. Transformer
c. Generator d. Thermo-couple
4. In Fleming's left-hand rule thumb always represents direction of
a. Current flow
b. Induced e.m.f
c. Magnetic field
d. Mechanical force
5.The field at any point on the axis of a current carrying coil will be
a. Perpendicular to the axis
b. Parallel to the axis
c. At an angle of 45° with the axis d. Zero
6. When a coil rotated in magnetic field the induced current in it
a. continuously changes
b. remains same
c. becomes zero
d. becomes maximum
7. In A.C generator increasing the number of turns in the coil.
a. decreases the electromotive force (e.m.f)
b. electromotive force (e.m.f) remains same c. increases the electromotive force (e.m.f)
d. electromotive force (e.m.f) becomes zero
8.The device in which induced e.m.f is statically induced e.m.fis
a. transformer
b. AC generator
d. dynamo c. alternator
9. What is the coefficient of mutual inductance, when the magnetic fluxchanges by 2x 10% Wb, and change in current is 0.01 A?
a. 2 H
b.3H
c. ½CH
d. zero
10. The induced e.m.f in a coil is proportional to
a. magnetic flux through the coil
b. rate of change of magnetic flux through the coil
c. area of the coil
d. product of magnetic flux and area of the coil
11. In a coil, current change from 2A to 4A in 0.05 s. If the average induced e.m.f is 8V then coefficient of self-induction is.
a. 0.2 henry
b. 0.1 henry
c. 0.8 henry
d. 0.04 henry
12. Which of the following quantities remain constant in step up transformer?
a. current
b. voltage
c.power
d. heat
13. Step-up transformer has a transmission ratio of 3:2. What is the voltage in secondary, if voltage in primary is 30 V?
a. 45 v
b. 15 v
c. 90 V
d. 300 V
14. Eddy current is produced when
a. a metal is kept in varying magnetic field
b. a metal is kept in steady magnetic field
c. circular coil is placed in a steady magnetic field
d. current is passed through a circular coil.
CONCEPTUAL QUESTIONS
Give a short response to the following questions:
1. What factors limit the size of the back e.m.f?
Ans. The magnitude of the back e.m.f induced in a coil depends upon magnetic
field strength and angular velocity. Explanation: Since back e.m.f is the induced e.m.f produced in a rotating coll olaced in magnetic field. The magnitude of this e.m.f is given by; E=NABWSinwt As the number of turns N and the cross-sectional area A of the coil remains con stant, therefore the magnitude of the back induced e.m.f depends upon magnetic field strength and angular velocity of the coil. Factors Affecting Back e.m.f: The factors which limit back e.m.fare;
1. Magnetic Field B: The greater the magnitude of magnetic field strength, the
Larger will be the magnitude of e.m.f produced due to changing flux.
2. Speed of Rotation of Armature: As the rate of change of flux depend upon the speed of the motor, therefore by increasing the speed of the motor willincrease the back e.m.f and decreasing its speed will decrease the back e.m.f. 3. Load: When load is connected to the motor, its speed decreases. Conse quently, the back e.m.f also decreases.
2. Why does back e.m.f tend to decrease as the rate of doing work increases?
Ans. By increasing rate of doing work, the angular velocity decreases which tend to decrease back e.m.f. Explanation: Consider a voltage source V connected to a motor which is rotating uniformly without load at a steady rate. The net potential which drives the current through the armature coil is given by;
V-E=IR
V=E+IR
Multiplying both sides, we get;
IV=18+FR ⇒ P=IE+FR
So the power supplied by the source is equal to the sum of power used to do work
against back e.m.f and the power dissipated in the coil. We see that when rate of doing work (12R) increases, the factor IE decreases. Cur rent is the common multiplier, therefore, the back e.m.f & will decrease.Reason: When the motor is connected with load then its rotational speed tends to decrease which reduces back e.m.f. NOTE: The decrease in the back e.m.f allows the motor to draw more current, i.e. V-E
1=R
3. In Faraday's experiments, what would be the advantage of using coils with many turns?
Ans. Using coils with many turns would make it easier to calculate the induced e.m.f and current.
Explanation: According to Faraday's law, induced e.m.f is directly proportional to the number of turns in the coil, i.e.Thus by increasing number of turns in a coil, greater induced e.m.f would be pro duced in the coil which result in greater current. It makes it easier to experimen tally measure these quantities. Many turns in the primary coil will make a larger magnetic flux, and many turns in the output coil will produce a larger output voltage and current.
4. Show that the relationship E=-40 is dimensionally correct.
Ans. The relationship will be dimensionally correct if both sides of the equation have the same dimensions.
Given that; mathcal E = - (Delta*phi)/(Delta*t)
Taking left hand side of the given equation, i.e. E
Since mathcal E = W/q = (Fs)/(it) = (mas)/(it) .
Here m = mass, a = i acceleration, s = displacement, I = current, t = time
Therefore in term of base units, the unit of & is given by; Unit of
mathcal C = (kgm * s ^ - 2 * m)/(As) = kgm'A's
Thus dimension of E is given by;
[ mathcal E ]= [M * L ^ 2 * A ^ - 1 * T ^ - 3] -> (1)
Now taking right hand side of the given equation, i.e.
(Delta*phi)/(Delta*t)
Since (Delta*Phi)/(Delta*t) = (Delta*BA)/(Delta*t) = (Delta * F/(ll) * A)/(Delta*t) = (Delta*maA)/(Delta*llt) = (maA)/(llt)
Here m = mass, a = acceleration, A = area, = current, L= length, t = time
(Delta*phi)/(Delta*t) Therefore in term of base units, the unit of is given by;
Unit of
ΔΦ _ kgms 'm
At
Ams
-=kgm'A's
Thus dimension of is given by;
(Delta*Phi)/(Delta*t)
[(Delta*Phi)/(Delta*t)] = [M * L ^ 2 * A ^ - 1 * T ^ - 3] -> (2)
By comparing Eq. (1) and Eq. (2), it is clear that the dimensions of both sides of the given relation are same. Hence the given relation is dimensionally correct.
5. Give the formulae for the flux linkage in terms of angular orientation.
Ans. The product of number of turns N in the coil and magnetic flux d linking the coil is called flux linkages. Formula: Consider magnetic field due to a coil of N turns in some region of space having area A. Then the flux linkage through the surface is given by;
Fluxlinkages-NO, (1) Since magnetic flux is the dot product of magnetic induction and area vector,therefore;
Q=B.A=BAcose
Putting this value in Eq. (1), we get;
Flux linkages=NBAcose →→→ (2) Thus Eq. (2) is represents the formula for the flux linkages in term of angular ori entation.
6. How electromagnetic brake works?
Ans. Brakes which work on the principle of electromagnetic induction are called electromagnetic brakes. It is also called eddy current brakes. Types: There are two types of electromagnetic brakes;
1. Linear EM brakes
2. Circular EM brakes
Working of Linear EM brakes: Linear EM brakes have two components; one is sta tionary and the other moves past it in a straight line. Usually, the stationary part is a heavy magnet and the other part is the object which we want to stop. As the moving object approaches the magnet (stationary), eddy currents are produced in it, i.e. in the moving object. These eddy currents have their own magnetic field. The two magnetic fields are in such a direction that their interaction makes the moving object stop.
Working of Circular EM brakes: Like linear electromagnetic brakes, circular EM brakes also have two parts; one static and the other moving. Further, in some cases, the electromagnetic part is static and the other is moving and in some other cases, the electromagnetic part is moving and the other is static.
In its simplest form, the electromagnet part is stationary and a metallic disc is ro tating inside it. When current passes through the electromagnet part, eddy cur rent passes through the electromagnet part, eddy currents are produced in the metallic disc which develops its own magnetic field. Again these magnetic fields interact and as a result the moving disccomes to a halt.
Advantage: Since EM brakes works on the principle of electromagnetic induction and there is no rubbing of surfaces. Therefore, the life time of the machine is in creased.
7. How eddy currents can be minimized in transformer?
Ans. Eddy Current: The induced currents that are set up in the core of transformer in the direction perpendicular to the flux are known as eddy currents. It results in power dissipation and heating of the core material.
Technique used to reduce Eddy Currents: In order to reduce the eddy currents losses and to improve efficiency of a transformer, the transformer cores are made of laminated iron, that is, many thin sheets of iron pressed together but separated by thin insulating layers. This limits the circulation of any eddy currents to the thickness of one lamina, rather than the whole core, thus reducing the overall heating effect.
Solid core with high Eddy currents.
Laminated core with low Eddy currents.
8. How electromagnetic induction is used in cook tops in electric ranges?
Ans. Cook tops in electric ranges works on the principle of electromagnetic induc tion producing eddy currents.
Explanation: Induction cookers use coils
(inductors) placed beneath a glass ce ramic cook top to generate heat for cook ing. Alternating current in the coils sets up an oscillating magnetic field which in duces eddy currents in metal pans placed in the vicinity of the varying field as shown in the figure. These currents cause the metal to get hot and hence heating its contents.
Limitation: It is important to mention that it is only useful for magnetic materials. Therefore, cooking utensils made up of aluminum, copper and non magnetic stain less steel is not compatible with an induction cook tops..
9. What is meant by the term back e.m.f in any electric motor operation? Why It is an advantage for the armature to rotate in a radial magnetic field rather than a uniform one?
Ans. Back e.m.f: The emf induced in the armature of a motor when it rotates in a changing magnetic field is called back emf in motors.
Explanation: When the coil motor rotates across the magnetic field by the applied potential difference V, an emf E is induced in it due to the changing magnetic flux. According to Lenz's law, the induced emf is in such a direction that opposes the emf running motor. Due to this reason, the induced emf is called back emf of the motor. The magnitude of back or counter e.m.f is proportional to the product of flux and the speed. Therefore magnitude of the back emf increases with the speed of motor.
Advantage of radial magnetic field: The advantage for the armature to rotate in a radial magnetic field rather than a uniform one is that, in radial magnetic field, the normal to the plane of the armature is always perpendicular to the magnetic field. Thus a maximum constant torque is acting on the coil which results in a constant acceleration of the armature.
10. Why it is an advantage for the armature to rotate in a radial magnetic field rather than a uniform one?
Ans. The advantage for the armature to rotate in a radial magnetic field rather than a uniform one is that in a radial magnetic field, uniform torque acts on the coil. Explanation: Since the torque acting on the coil in a magnetic field is given by: T= NIBA sino In case of radial field, normal to the plane of coil is always perpendicular to the magnetic field, i.e. 0-90° Therefore, the torque acting on the coil will be:
NIBA sin90°
TNIBA
sin90=1
Thus in radial field, the torque acting on the coil is maximum and uniform.
11. Which materials would you prefer while making transformer core; iron, solid soft iron, laminated soft iron, aluminum? (b). what is the reason for using this material?
Ans. Main features of transformer core: The use of transformer core is to concen trate the strength and increase the effect of magnetic fields produced by electric currents. In selecting a transformer core material, it is desired that the core should not retain magnetization when the applied field is removed, this property, called hysteresis can cause energy losses in transformers. The core makes up the bulk of a transformer, so it's no surprise that selecting the proper material plays an integral role in the transformer's overall function. (a) Since aluminum is non-magnetic, therefore it cannot be used as transformer core. Also iron and solid soft iron can be used as a transformer core, but due to high energy losses as eddy currents and hysteresis, these are not pre ferred. We would like to use laminated soft iron for transformer core. (b) Soft iron has smaller hysteresis loss and laminated iron effectively reduces eddy current than solid soft iron. This is because of thins insulated sheets.
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