This set of Physics Multiple Choice Questions & Answers (MCQs) focuses on “Electromagnetic Induction – Motional Electromotive Force”.

1. Identify the expression for the motional electromotive force from the following?

a) E = -vLB

b) E = vLB

c) E = \(\frac {v}{LB}\)

d) E = \(\frac {LB}{v}\)

View Answer

Explanation: Motional electromotive is the emf induced by the motion of the conductor across the magnetic field. The expression for motional electromotive force is given by:

E = -vLB

This equation is true as long as the velocity, magnetic field, and length are mutually perpendicular to each other. The negative sign is associated with Lenz’s law.

2. A bar of length 0.7 m slides along metal rails at a speed of 1 m/s. The bar and rails are in a magnetic field of 20 T, pointing out into the page. Calculate the motional emf.

a) 0.7 V

b) 7 V

c) 14 V

d) 1.4 V

View Answer

Explanation: Length (L) = 0.7 m; Speed (v) = 1 m/s; Magnetic field (B) = 20 T

The required equation E = -vLB (The negative sign only applies to the direction)

E = 1 × 0.7 × 20

E = 0.7 × 20

E = 14 V

Therefore, the motional emf in the bar and rails is 14 V.

3. A bar of length 0.15 m slides along metal rails at a speed of 5 m/s. The bar and rails are in a magnetic field of 40 T, pointing out into the page. The resistance of two resistors in parallel is both 20 Ω, and the resistance of the bar is 5 Ω. What is the current in the bar?

a) 1 A

b) 2 A

c) 3 A

d) 5 A

View Answer

Explanation: L = 0.15 m; B = 40 T; v = 5 m/s; R1 = 10 Ω; R2 = 10 Ω; R3 = 5 Ω

Emf (E) = vLB = 5 × 0.15 × 40

E = 5 × 6

E = 30 V

R1 and R2 are in parallel ➔ \(\frac {1}{R} = \frac {1}{R1} + \frac {1}{R2} = \frac {1}{20} + \frac {1}{20} = \frac {2}{20} = \frac {1}{10}\) ➔ R = 10 Ω

R

_{TOT}= R + R3 = 10 + 5 = 15 Ω

Therefore, current (I) = \(\frac {E}{R}\)

I = \(\frac {30}{15}\)

**I = 2 A**

4. Induced emf and motional emf are exactly the same.

a) True

b) False

View Answer

Explanation: No. they are not exactly the same. According to Faraday’s Law, an induced emf is created whenever there’s a changing magnetic flux through a loop. If the changing emf is due to some kind motion of a conductor in a magnetic field, then it would be called as motional emf.

5. A bar of length 2m is said to fall freely in a magnetic field of magnitude 50 T. What is the motional emf in the bar when it has fallen 40 meters?

a) 700 V

b) 2100 V

c) 2800 V

d) 1400 V

View Answer

Explanation: Given: L = 2 m; B = 50 T;

The bar is said to be falling freely, so ➔ u = 0

According to Newton’s third equation ➔ v

^{2}– u

^{2}= 2gh

v

^{2}= 2gh; h = 40 m

v = \(\sqrt{(2 × 9.8 × 40)}\)

v = 28 m/s

Motional emf (E) = vLB

E = 28 × 2 × 50

E = 2800 V

6. A metal rod is forced to move with constant velocity along two parallel metal rails, connected with a strip of metal at one end across a magnetic field (B) of 0.5 T, pointing out of the page. The rod is of length 45 cm and the speed of the rod is 70 cm/s. The rod has a resistance of 10 Ω and the resistance of the rails and connector is negligible. What is the rate at which energy is being transferred to thermal energy?

a) 0.225 W

b) 22.55 W

c) 2.25 × 10^{-4} W

d) 2.25 × 10^{-3} W

View Answer

Explanation: Given: B = 0.5 T; v = 70 cm/s = 70 × 10

^{-2}m/s; L = 45 cm = 45 × 10

^{-2}m; R = 10 Ω

Motional emf (E) = vLB = 70 × 10

^{-2}× 45 × 10

^{-2}× 0.5

E = 0.15 V

Current (I) = \(\frac {E}{R}\)

I = \(\frac {0.15}{10}\)

I = 0.015 A

Rate of energy or power (P) = I

^{2}R

P = 0.015

^{2}× 10

P = 2.25 × 10

^{-3}W

Therefore, the rate of energy transfer is 2.25 × 10

^{-3}W.

7. Find the true statement.

a) Motional emf is inversely proportional to speed of electric conductor

b) Motional emf is inversely proportion to the length of the conductor

c) Motional emf is directly proportional to the magnetic field

d) Motional emf is inversely proportional to the magnetic field

View Answer

Explanation: Motional emf is directly proportional to the magnetic field across which the electric conductor moves. It is also directly proportional to the \(\frac {velocity}{speed}\) of the electric conductor as well as to the length of the conductor. All the other statements are not valid.

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