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A metal rod $O A$ of mass $m$ and length $r$ is kept rotating with a constant angular speeds $ω$ with a uniform magnetic field $B$ applied in horizontal direction. An inductor $L$ of resistance $R$ are connected through a switch $S$ across points $O$ and $C$ on the ring over which rod's free end is rotating in vertical plane.
Neglect the resistance of the rod, ring and connecting wires. Obtain the value of the torque that is required to maintain the constant speed of the rotating rod. When current in the wire reaches its steady state. At $t = 0$ , switch is closed.

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$τ = \frac{m g r \cos θ}{2} + \frac{1}{4} \frac{B^{2} r^{4} ω}{R}$

$τ = \frac{1}{4} \frac{B^{2} r^{4}}{R} + m g R$

$τ = m g R \sin θ + B r$

$τ = m g r \cos θ$

answer is A.

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Detailed Solution

Induced emf, $ε = \frac{1}{2} B r^{2} ω$

Under steady state

$i = \frac{ε}{R} = \frac{1}{2} (\frac{B r^{2} ω}{R})$

Force on rod due to this induced current,

$F = B I L = B \times \frac{1}{2} \frac{B r^{2} ω}{R} \times r = \frac{1}{2} \frac{B^{2} r^{3} ω}{R}$

Torque due to this force $= F \times$ distance

$= F \times \frac{r}{2} = \frac{1}{4} . \frac{B^{2} r^{4} ω}{R}$

Torque due to weight of rod at $θ$ angular position

$= m g \cos θ \times \frac{r}{2}$

So, total torque required $= \frac{m g r \cos θ}{2} + \frac{1}{4} \frac{B^{2} r^{4} ω}{R}$

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