JEE MAIN - Physics (2004 - No. 36)
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
$$5mV$$
$$50\mu V$$
$$5\mu V$$
$$50mV$$
Explanation
$$\ell = 1m,\,\,\omega = 5\,rad/s,\,\,B = 0.2 \times {10^{ - 4}}T$$
$$\varepsilon = {{B\omega {\ell ^2}} \over 2} = {{0.2 \times {{10}^{ - 4}} \times 5 \times 1} \over 2} = 50\mu V$$
$$\varepsilon = {{B\omega {\ell ^2}} \over 2} = {{0.2 \times {{10}^{ - 4}} \times 5 \times 1} \over 2} = 50\mu V$$
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