A rod of length 1 m rotates at 60 rad/s about one end in a field of 0.2 T parallel to the axis of rotation. The emf between its ends is
Answer: B
The emf is \( \tfrac{1}{2}B\omega l^2 \), that is \( 0.5 \times 0.2 \times 60 \times 1 = 6 \) V. The factor of one half arises because the velocity of a point on the rod grows linearly from zero at the axis, so its average is half the tip value.
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