JEE Main · 2004 · 24 Apr 2004 · Offline · Electromagnetic Induction

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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 1 0^{- 4} T

, then the e.m.f. developed between the two ends of the conductor is

(A) $5 μ V$
(B) $50 μ V$ ✓ Correct
(C) $5 mV$
(D) $50 mV$

Answer: (B)

Solution

A straight rod turning about one of its ends cuts the perpendicular field uniformly, and the emf set up between its ends is

ε = \frac{1}{2} B ω L^{2}

, with

B

the component of the field at right angles to the plane of rotation. Putting the horizontal component

B = 0.2 \times 1 0^{- 4} T

ω = 5 rad/s

and

L = 1 m

gives

ε = \frac{1}{2} \times 0.2 \times 1 0^{- 4} \times 5 \times 1^{2} = 0.5 \times 1 0^{- 4} V = 50 μ V

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option

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JEE Main · 2004 · 24 Apr 2004 · Offline · Electromagnetic Induction

Solution

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