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Q. A wheel with $10$ metallic spokes each $0.5 \,m$ long is rotated with a speed of $120 \,rev / min$ in a plane normal to the earth's magnetic field at the place. If the magnitude of the field is $0.4 \,G$, the induced e.m.f between the axle and the rim of the wheel equal to

Solution:

$e=\frac{1}{2} B \omega l^{2}$
$=\frac{1}{2} \times 0.4 \times 10^{-4} \times 2 \pi \times \frac{120}{60} \times \frac{1}{4}$
$=2 \times 3.14 \times 10^{-5}$
$=6.28 \times 10^{-5} V$