Electromagnetic Induction Multiple Choice Questions And Answers
Question 1. Lenz’s law is the consequence of the law of conservation of:
- Energy
- Charge
- Mass
- Momentum
Answer: 1. Energy
Question 2. The emf induced in a 10H inductor in which current changes from 11 A to 2A in 9 x 10-1 s is :
- 104 V
- 103 V
- 102 V
- 10 V
Answer: 3. 102 V
⇒ \(\mathrm{e}=-\mathrm{L} \frac{\Delta \mathrm{I}}{\Delta \mathrm{t}}\)
∴ \(e=\frac{-10 \times(2-11)}{9 \times 10^{-1}}=10^2 \mathrm{~V}\)
Question 3. A metal plate is getting heated. Which one of the following statements is incorrect?
- It is placed in a space-varying magnetic field that does not vary with time.
- A direct current is passing through the plate.
- An alternating current is passing through the plate.
- It is placed in a time-varying magnetic field.
Answer: 1. It is placed in a space-varying magnetic field that does not vary with time.
Question 4. The magnetic flux linked with a coil is given by Φ = 5t2+ 3t + 10, where Φ is in Weber and l is in second. The induced emf in the coil at t = 5 sec will be
- 53 V
- 43 V
- 10 V
- 6 V
Answer: 1. 53 V
⇒ \(\mathrm{e}=-\frac{\mathrm{d} \phi}{\mathrm{dt}}\)
e = 10t + 3
e = 10(5) + 3 = 53V
Question 5. The magnetic flux linked with a coil changes with time t (second) according to Φ = 6t2– 5t + 1, where Φ is in Wb. At l = 0.5 S, the induced current in the coil is ______ The resistance of the circuit is 10 Ω.
- 1 A
- 0.1 A
- 0.1 mA
- 10 A
Answer: 2. 0.1 A
⇒ \(\mathrm{e}=-\frac{\mathrm{d} \phi}{\mathrm{dt}}\)
e = -(12t-5)
e = -(12 x 0.5-5)
e = 1
⇒ \(|\mathrm{e}|=1 \mathrm{~V}\)
∴ \(I=\frac{e}{R}=\frac{1}{10}=0.1 \mathrm{~A}\)
Question 6. A square conducting coil of area 100 cm2 is placed normally inside a uniform magnetic field of 103 Wbm-2. The magnetic flux linked with the coil is ______ wb.
- 10
- 10-5
- 10-g
- 0
Answer: 1. 10
= BA
= 103 x 100 x 10– 4
= 100 weber
Question 7. The self-inductance of two solenoids A and B having equal lengths are the same. If the number of turns in two solenoids A and B arc 100 and 200 respectively. The ratio of radii of their cross-section will be
- 2:1
- 1:2
- 1:4
- 4:1
Answer: 1. 2:1
⇒ \(L=\frac{\mu_r \mu_0 N^2 \pi r^2}{l}\)
⇒ \(r^2=\frac{L \times l}{\mu_r \mu_0 N^2 \pi}\)
l, L, r, 0, – constant
\(r \propto \frac{1}{N}\) \(\frac{r_1}{r_2}=\frac{N_2}{N_1}=\frac{200}{100}=2: 1\)Question 8. A magnet is moving towards a coil along its axis and the emf induced in the coil is s. If the coil also starts moving towards the magnet with the same speed, the induced emf will be
\(\frac{\varepsilon}{2}\)2
4
Answer: 2. 2
Question 9. The dimensional formula of self-inductance is __________.
M1L2T-2A-2
M1L1T2A-2
M1L1T-2A-2
M1L1T-1A-1
Answer: 1.
Question 10. The mutual inductance of the system of two coils is 5 mH. The current in the first coil varies according to the equation T = I0 sin cot, where I0 = 10A and co = 100 TT rad/s. The value of maximum induced emf in the second coil is __________.
2V
5V
V
4V
Answer: 2. 5V
\(\mathrm{e}=-\mathrm{M} \frac{\mathrm{dI}}{\mathrm{dt}}\)= -M(I0 cos t) [for max. value cos t = 1]
= -5 x 10-3 x 10 x 100 x (1)
e = -5V
|e| = 5V
Question 11. Current of 2A passing through a coil of 100 turns gives rise to a magnetic flux of 5 x 10-3 Wb per turn. The magnetic energy associated with coil is __________.
5 x 10-3 J
0.5 x 10-3
5 J
0.5 J
Answer: 4. 0.5 J
\(\mathrm{u}=\frac{1}{2} \mathrm{LI}^2 \quad \phi=\mathrm{LI}\)= \(\frac{1}{2} \mathrm{\phi}{I}=\frac{1}{2} \times 5 \times 10^{-3} \times 100 \times 2=0.5 \mathrm{~J}\)
Question 12. The flux linked per turn of a coil of N turns changes from (|)1 and <f)2- If die total resistance ol the circuit including the coil is R, the induced charge in the coil.
\(-N \frac{\left(\phi_2-\phi_1\right)}{l}\) \(-N \frac{\left(\phi_2-\phi_1\right)}{R}\) \(-N \frac{\left(\phi_2-\phi_1\right)}{R t}\) \(-\mathrm{N}\left(\phi_2-\phi_1\right)\)Answer: 2. \(-N \frac{\left(\phi_2-\phi_1\right)}{R}\)
\(q=-\frac{N}{R} \Delta \phi\) \(q=-\frac{N\left(\phi_2-\phi_1\right)}{R}\)Question 13. The magnitude of the induced cmf is equal to the lime rate of change of ____.
Electric flux
Magnetic force
Magnetic flux
Electric force
Answer: 3. Magnetic flux
Question 14. Unit of induced emf is _____.
Tesla
Volt/second
weber/second
volt
Answer: 4. Volt
Question 15. One conducting wire of length 50 cm is moving perpendicular to uniform magnetic field of 0.2 T. with constant velocity of 10 ms 1 cmf induced between two ends of a wire is _________ V.
1.0
0.1
0.01
10
Answer: 1. 1.0
\(e=-10 \times 0.2 \times \frac{50}{100}\)e = -1 Volt
|e| = 1 V
Electromagnetic Induction Assertion And Reason
For question numbers 1 to 5 two statements arc given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (1), (2), (3) and (4) as given below.
- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is NOT the correct explanation of A
- A is true but R is false
- A is false and R is also false
Question 1. Assertion: Induced cmf in two coils made of wire of the same length and thickness, one of copper and another of aluminium is the same. The current in copper coil is more than the aluminium coil.
Reason: Resistance of aluminium coil is more than that of copper coil.
Answer: 1. Both A and R are true and R is the correct explanation of A
Question 2. Assertion: If the current in an inductor is zero at any instant, then the induced emf may not be zero.
Reason: An inductor keeps the flux (i.e. current) constant.
Answer: 1. Both A and R are true and R is the correct explanation of A
Question 3. Assertion: Self-induction is called the inertia of electricity.
Reason: The inductor opposes the change in current.
Answer: 1. Both A and R are true and R is the correct explanation of A
Question 4. Assertion: A small magnet takes a longer time to fall into a hollow metallic tube without touching the wall.
Reason: There is an opposition of motion due to the production of eddy currents in metallic tubes.
Answer: 1. Both A and R are true and R is the correct explanation of A
Question 5. Assertion: Faraday’s laws are consequences of the conservation of energy.
Reason: In a purely resistive AC circuit, the current lags behind the EMF in phase.
Answer: 4. A is false and R is also false
Electromagnetic Induction Short Questions And Answers
Question 1. Two identical circular discs, one of copper and another of aluminium, are rotated about their geometrical axes with the same angular speed in the same magnetic field acting perpendicular to their planes. Compare the (1) induced cmf, and (2) induced current produced in discs between their centre and edge. Justify your answers,
Answer:
Emf will be same for both, because it depends on N, A, B, ro and that arc same for both disc But induced current \(I=\frac{\mathrm{e}}{\mathrm{R}}=\frac{\mathrm{NAB}\omega}{\mathrm{R}}\)
Copper has less resistance so induced current will be higher in copper.
Question 2. A rectangular loop PQMN with movable arm PQ of length 10 cm and resistance 2£2 is placed in a uniform magnetic field of 0.1 T acting perpendicular to the plane of the loop as is shown in the figure. The resistances of the arms MN, NP and MQ are negligible. Calculate the
- emf induced in the arm PQ and
- Current induced in the loop when arm PQ is moved with velocity 20 m/s
Answer:
1. Induced emf in arm PQ
e = -B/v
c = -0.1 x 10 x 1 0-2 x 20
e = -0.2 V
2. Current induced in loop
\(I=\frac{|e|}{R}=\frac{0.2}{2}=0.1 \mathrm{~A}\)Question 3. Predict the polarity of the capacitor in the situation described below:
image
Answer:
Plate A (+ve) and plate B(-ve)
Question 3. An aeroplane is flying horizontally from west with a velocity of 900 km/hour. Calculate the potential difference developed between the ends of its wings having a span of 20m. The horizontal component of the Earth’s magnetic field is 5 x Hr4 T and the angle of dip is 30°.
Answer:
Potential difference developed between the ends of the wings ‘e’ = Blv
Given Velocity (v) = 900 km/hour
= 250 m/s
Wing span (l) = 20 m
Vertical component of Earth’s magnetic field
Bv = BH tan (5
= 5 x 10-4 (tan 30°) tesla
∴ Potential Difference
e = BVl
e = 5 x 10-4 (tan 30°) x 20 x 250
\(e=\frac{5 \times 20 \times 250 \times 10^{-4}}{\sqrt{3}}\)e = 1 .44 volt
- Question 4.
- Define self-inductance. Write its SI unit.
- A long solenoid with 15 turns per cm has a small loop of area 2.0 cm2 placed inside the solenoid normal to its axis. If the current canned by the solenoid changes steadily from 2.0 A to 4.0 A in 0. 1 s. what is the induced cmf in the loop while the current is changing?
Answer:
1. Self inductance of a coil is numerically equal to cmf induced in the coil, when rate of
change of current is unity.
Or
L = N<(>/I
It is equal to the magnetic flux linked with coil when a unit current flow through it.
The S.I unit of L is henry (H).
2. Here, number of turns per unit length.
n= N/l = 15 turns/cm = 1500 turns/m
A = 2 cm2 = 2 x 10-4 m2
dl/dt = (4- 2)/0. 1 or dl/dt = 20 As-1
⇒ \(|\mathrm{e}|=\frac{\mathrm{d} \phi}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{BA})\) (∵ \(B=\frac{\mu_0 N I}{l}\) )
⇒ \(|\mathrm{e}|=\frac{\mathrm{Ad}}{\mathrm{dt}}\left(\mu_0 \frac{\mathrm{NI}}{l}\right)=\mathrm{A} \mu_0\left(\frac{\mathrm{N}}{l}\right) \frac{\mathrm{dI}}{\mathrm{dt}}\)
|e| = (2 x 10-4) x 4 x 10-7 x 1500 x 20V
|e| = 7.5 x 10-6 V
Question 5.
- Define mutual inductance.
- A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20A in 0.5 s. what is the change of flux linked with the other coil?
Answer:
1. Mutual Inductance : It is numerically equal to the magnetic flux linked with one coil (secondary coil) when a unit current flows through the other coil (primary coil).
2. Magnetic flux
initial = MI initial = 0 as I initial = 0
and final = Mfinal = 1.5 x 20 = 30Wb
Change in flux d= 1-2 = 30-0 = 30 weber.
Question 6. Define the self-inductance of a coil. Obtain the expression for the energy stored in an inductor L connected across a source of cmf.
Answer:
It is numerically equal to the induced emf per unit rate of change of current through a given coil.
i.e. \(\mathrm{L}=\frac{\mathrm{e}}{-\frac{\mathrm{dI}}{\mathrm{dt}}}\)
Energy stored in an inductor: Work has to be done by battery against the opposing induced cmf in establishing a current in an inductor. The energy supplied by the battery is stored in the inductor.
Let current flowing through the circuit at any instant (t) be i.
Rate of change of current at that time = \(\frac{\mathrm{di}}{\mathrm{dt}}\)
image
magnitude of induced emf in the inductor,
\(\mathrm{e}=\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}\)Work done by the battery in time dt,
dW = Vdq = edq
dW = eidt
\(d W=\left(L \frac{d i}{d t}\right) i d t\)dW = Lidi
Total work done by the battery in increasing current from 0 to I
\(\mathrm{W}=\mathrm{L} \quad \int \mathrm{i} d i \Rightarrow \mathrm{W}=\frac{1}{2} \mathrm{LI}^2\)This work done is stored in an inductor in the form of magnetic energy.
\(\mathrm{U}_{\mathrm{B}}=\frac{1}{2} \mathrm{LI}^2\)Question 7. Define mutual inductance between a pair of coils. Derive an expression for the mutual inductance of two long coaxial solenoids of the same length wound one over the other.
[OR. prove that M12 = M21]
Answer:
Mutual Inductance: It is numerically equal to the magnetic flux linked with one coil (secondary coil) when unit current flows through the other coil.
Derivation:
image
For I coil, <|)1 = M12, I2 (1)
where <j)1 is the flux through I coil.
Similarly,
1 = (B2.A1)N1
= \(\)
= \(\) (2)
From (1) and (2)
\(\)Now, 2 = M21I1 (3)
where ())2 is the flux through 2nd coil due to Ist coil.
Also. <|)-, = B1 A1 N2 cos0
= \(\) (4)
(0 = 0°)
From (3)and (4)
\(\) \(\)∴ M12 = M21
Electromagnetic Induction Long Questions And Answers
Question 1. The emf induced across the ends of a conductor due lo its motion in a magnetic field is called motional emf. It is produced due to the magnetic Lorcnlz force acting on the free electrons of the conductor. For a circuit shown in figure, if a conductor of length f moves with velocity v in a magnetic field B perpendicular to both its length and the direction of the magnetic field, then all the induced parameters are possible in the circuit.
image
(1). Which rule is used to find direction of induced current?
Answer:
Fleming’s right hand rule
(2). Bicycle generator creates 1 .5 V at 15 km/hr. Find the value of EMF generated at 10 km/hr ?
Answer:
e = NABo
e
\(\frac{1.5}{15}=\frac{x}{10}\)x = 1 volt.
(3). A O.i m long conductor carrying a current of 50 A is held perpendicular to a magnetic field of 1.25 mT. Find the mechanical power required to move the conductor with a speed of 1 m s-1 ?
Answer:
P = F x v = (IlB)v = 50 x 0.1 x 1.25 x 10-3 x 1
= 6.25 x 10-3 W = 6.25 mW
(4). A conducting rod of length f is moving in a transverse magnetic field of strength B with velocity v. The resistance of the rod is R. Find the expression of induced current in the rod?
Answer:
\(I=\frac{e}{R}=\frac{B v l}{R}\)Question 2. Mutual inductance is the phenomenon of inducing cmf in a coil, due to a change of current in the neighboring coil. The amount of mutual inductance that Jinks one coil to another depends very much on the relative positioning of the two coils, their geometry and relative separation between them. Mutual inductance between the two coils increases pr, limes if the coils arc wound over an iron core of relative permeability pr.
image
(1). A short solenoid of radius a, number of turns per unit length n, and length L is kept coaxially inside a very long solenoid of radius b, number of turns per unit length n2. Write the expression for mutual inductance of the system?
Answer:
p07ta2n,n2L
(2). If a change in current of 0.01 A in one coil produces a change in magnetic flux of 2 x 10-2 Weber in another coil. Find the mutual inductance between coils?
Answer:
= MI
\(M=\frac{\phi}{I}=\frac{2 \times 10^{-2}}{0.01}\)M = 2H
(3). How can mutual inductance between two coils can be increased?
Answer:
By increasing the number of turns in the coils
\(M=\frac{\mu_r \mu_0 N^2 A}{l}\)M N2
(4). When a sheet of iron is placed in between the two co-axial coils. How the mutual inductance between the coils will change?
Answer:
Increases
\(\mathrm{M}=\frac{\mu_{\mathrm{r}} \mu_0 \mathrm{~N}^2 \mathrm{~A}}{l}\)M r
Question 3. Explain the meaning of the term mutual inductance. Consider two concentric circular coils. one of radius R and the other of radius r (R > r) placed coaxially with centres coinciding with each other. Obtain the expression for the mutual inductance of the arrangement.
A rectangular coil of area A, having number of turns N is rotated at T revolutions per second in a uniform magnetic field B, the field being perpendicular to the coil. Prove that the maximum emf induced in the coil is 2 7tf NBA.
Answer:
1. Let current (I1) is flowing in first outer coil. The magnetic field at centre of coil
\(B_1=\frac{\mu_0 I_1}{2 R}\)Flux passing through second inner coil
2 = B1 A2
\(\phi_2=\frac{\mu_0 I_1}{2 \mathrm{R}} \times \pi \mathrm{r}^2\)From definition of mutual induction
2 = M21 I1
\(M_{21}=\frac{\phi_2}{I_1}=\frac{\mu_0 \pi r^2}{2 R}\) \(M_{21}=\frac{\mu_0 \pi r^2}{2 R}\)Similarly \(M_{12}=\frac{\mu_0 \pi r^2}{2 R}\)
M12 = M21 = M
\(M=\frac{\mu_0 \pi r^2}{2 R}\)2. Magnetic flux linked with coil = <j) = NBAcosG
The induced cmf.e = -d(|)/dt (∵ \(\omega =\frac{\mathrm{d} \theta}{\mathrm{dt}}\\) )
\(\mathrm{e}=-\frac{\mathrm{d} \phi}{\mathrm{dt}}=\left(-\mathrm{NBA}(-\sin {\theta}) \frac{\mathrm{d} \theta}{\mathrm{dt}}\right)=\mathrm{NBA} \sin {\theta}\left(2 \pi \mathrm{f}\right)\)For maximum induced emf, sin0 = 1
∴ e = NBA(2f) => e = 27ilNBA
Question 4.
Draw a labelled diagram of an ac generator. Obtain the expression for the cmf induced in the rotating coil of N turns each of cross-sectional area A, in the presence of a magnetic field B.
A horizontal conducting rod 10 m long extending from cast to west is falling with a speed
5.0 ms-1 at right angles to the horizontal component of the Earth’s magnetic field,
0.3 x 10-4 Wb m-2. Find the instantaneous value of the emf induced in the rod.
Answer:
1. image
B1 and B2 are carbon brushes. S1 and S2 are slip rings.
Let at any instant (t), the angle between \(\vec{B}\) and \(\vec{A}\) is 0, then the magnetic flux linked with coil at that instant.
\(\phi_{\mathrm{B}}=N(\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{A}})\)= NBA cos
N = Number of turns in the coil
A = Area of the coil
0) = Angular velocity of coil
[ ∵ 0 = rot ]
cj)B = NBA cos rot
\(\frac{d \phi_{\mathrm{B}}}{\mathrm{dt}}=\mathrm{NBA} \omega(-\sin \omega t)\)Thus induced emf, \(\mathrm{e}=-\frac{\mathrm{d} \phi_{B}}{\mathrm{dt}}\)
or e = NBA sint [ ∵ = t]
2. Given: L = 10 m, v = 5 m/s, BH = 0.3 x 10-4 Weber
Induced emf (e) = BH vL
= 0.3 x 10-4 x 5 x 10
= 15 x 10-4 volt
= 1.5 mV
Question 5.
When a bar magnet is pushed towards (or away) from the coil connected to a galvanometer. the pointer in the galvanometer deflects. Identify the phenomenon causing this deflection and write the factors on which the amount and direction of the deflection depends. Slate the laws describing this phenomenon.
Sketch the change in flux and emf when a conducting rod PQ of resistance R and length l moves freely to and fro between A and C with speed v on a rectangular conductor placed in uniform magnetic field as shown in the figure.
image
Answer:
1. When a bar magnet pushed towards or away from coil, magnetic flux passing through coil
change with lime and causes an induced emf. This phenomenon is called EMI.
Induced emf in the coil is given as
\(\epsilon =-\mathrm{N} \frac{\mathrm{d}\phi}{\mathrm{dt}}\)Direction and amount of deflection depend on the motion of magnet whether it is moving
towards the coil or away from the coil and the speed with which it moves.
Farady’s law of Electromagnetic Induction:
Whenever there is a change in magnetic flux linked with a coil, an emf is induced in the coil. The induced emf is proportional to the rate of change of magnetic flux linked with the coil.
\(\varepsilon \propto \frac{\mathrm{d} \phi}{\mathrm{dt}}\) \(\varepsilon \propto \frac{-\mathrm{d} \phi}{\mathrm{dt}} \Rightarrow \varepsilon=-\mathrm{K} \frac{\mathrm{d} \phi}{\mathrm{dt}}\)Where K is constant and negative sign represents oppositions to change in flux. In SI system is in weber, t in seconds, in volt and K = 1
∴ \(\varepsilon=-\frac{\mathrm{d} \phi}{\mathrm{dt}}\)
If the coil has N turns, then
\(\varepsilon=-\mathrm{N} \frac{\mathrm{d} \phi}{\mathrm{dt}}\)2. Case 1 : When PQ moves forward
(1). For 0 x < b
Magnetic field, B exists in the region
∴ Area of loop PQRS = l x
∴ Magnetic flux linked with loop PQRS.
<]) = BA = Blx
(|) = Blx (1) [b > x 0]
(2). For 2b > x > b
B =0
∴ Flux linked with loop PQRS is uniform and given by
‘ = Blb (x = b)
Forward jounery
Thus, for 2b x bar
Flux, = Bbl [constant]
Return journey
For b x 2b
Flux, = constant = Bbl [Decreasing]
Graphical representation
case – 2 For b> x 0
As = Blx = \(\frac{\mathrm{d} \phi}{\mathrm{dt}}=\frac{\mathrm{Bl}\mathrm{dx}}{\mathrm{dt}}=\mathrm{Bv} l\) ( ∵ \mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}\(\))
Induced emf, e = \(\mathrm{c}=-\mathrm{d} \phi / \mathrm{d} \mathrm{l}=-\mathrm{vB}\left(\Rightarrow \frac{\mathrm{d} \phi^{\prime}}{\mathrm{dt}}=0\right)\)
For 2b x b
As = bb => c = 0
Forward journey,
For b > x 0
c = -vbl,
For 2b x b.e= 0
Return journet,
For b > x 0
e= vBl
For 2b x n2e=0
image