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A metallic ring of radius r with a uniform metallic spoke of negligible mass and length r is rotated about its axis with angular velocity $ω$ in a perpendicular uniform magnetic field B as shown in figure. The central end of the spoke is connected to the rim of the wheel through a resistor R as shown. The resistor does not rotate, its one end is always at the center of the ring and the other end is always in contact with the ring. A force F as shown is needed to maintain constant angular velocity of the wheel. F is equal to (the ring and the spoke has zero resistance)

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$\frac{B^{2} ω r^{2}}{8 R}$

$\frac{B^{2} ω r^{2}}{2 R}$

$\frac{B^{2} ω r^{3}}{2 R}$

$\frac{B^{2} ω r^{3}}{4 R}$

answer is D.

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Detailed Solution

Induced emf in the spoke is shown in figure below.

$e = \frac{1}{2} {Bωr}^{2}, I = \frac{e}{R} = \frac{{Bωr}^{2}}{2 R}$

There will be no induced emf separately in parts ADC or AEC.

$F_{b} = IrB = \frac{B^{2} {ωr}^{3}}{2 R}$

Balancing torque about E: $Fr = \frac{F_{b} r}{2}$

$\Rightarrow F = \frac{F_{b}}{2} = \frac{B^{2} {ωr}^{3}}{4 R}$

Note: Force due to currents $(\frac{I}{2})$ will act on circular parts also, but their torque about E will be zero.

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