In the figure shown R is the 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 clock wise direction with a uniform angular velocity $ω$ . There is a uniform magnetic field of strength ‘B’ pointing inwards. $r$ is a stationary resistance.

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

Direction of current in external r is from center to circumference.

Current through r is zero

Direction of current in external r is from circumference to center.

answer is B, D.

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Detailed Solution

Equivalent circuit :-

Induced emf $e = \frac{B ω r^{2}}{2} e = \frac{B ω a^{2}}{2}$

By nodal equation :-
$4 (\frac{X - e}{r}) + (\frac{X - o}{r}) = 0 X = \frac{4 e}{5} = \frac{2 B w a^{2}}{5} a n d I = \frac{x}{r} = \frac{2 B w a^{2}}{5 r}$
Also direction of current in r will be from rim to origin. (Towards negative terminal)