First slide

Motional EMF

Question

A metal conductor of length 1 m rotates vertically about one of its ends at angular velocity 5 radians per second. If the horizontal component of earth's magnetic field is $0.2 \times 10^{- 4} T$ then the e.m.f. developed between the two ends of the conductor is

Easy

A

5 mV
B

$5 \times 10 V$
C

50 mV
D

$50 μ V$

Solution

$e = \frac{1}{2} {Bωr}^{2} = \frac{1}{2} \times 0 .2 \times 10^{- 4} \times 5 \times (1)^{2} = 50 μV$

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