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In figure. M is fixed conducting ring of negligible resistance and radius ‘ $a$ ’. PQ is a uniform rod of resistance ‘ $r$ ’. It is hinged at the centre of the ring and rotated about this point in clockwise direction with a uniform angular velocity $ω .$ There is a uniform magnetic field of strength B pointing inward and ‘ $r$ ’ is a stationary resistance. Then

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Current through ‘ $r$ ’ is zero

Current through ‘ $r$ ’ is $\frac{2 B ω a^{2}}{5 r}$

Direction of current in external resistance ‘ $r$ ’ is form centre to circumference

Direction of current in external resistance ‘ $r$ ’ is from circumference to center

answer is B, D.

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Detailed Solution

Equivalent circuit :

Indeed e.m.f, $E = \frac{B ω R^{2}}{2} = (\frac{B ω a^{2}}{2}) (∵ r a d i u s = a)$

Also direction of current in (i) will be toward negative terminal, i.e., from rim to origin alternating, by equivalent of cells figure $I = \frac{E}{r + \frac{r}{4}} = \frac{4 E}{5 r} = \frac{2 B ω a^{2}}{5 r}$

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