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A line charge $λ$ per unit length is lodged uniformly onto the rim of a wheel of mass M and radius R. The wheel has light non-conducting spokes and is free to rotate without friction about its axis (see figure) . A uniform magnetic field extends over a circular region within the rim . It is given by, $\vec{B} = - B_{0} \hat{k} (r \leq a, a < R)$ $\vec{B} = 0$ otherwise.
What is the angular velocity of the wheel after the field is suddenly switched off?

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$ω= \frac{{Bπa}^{3} λ}{MR}$

$ω= \frac{{Bπ}^{2} a^{2} λ}{MR}$

$ω= \frac{{Bπa}^{2} λ}{MR}$

$ω= \frac{{Bπ}^{2} a^{2} λ^{2}}{M^{2} R^{2}}$

answer is B.

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Detailed Solution

Electric field induces at each point of the rim. Which exerts tangential force on the rim. This force constitutes a torque, and so rim starts rotating.

we know that, $e = \oint \vec{E} \cdot d \vec{l} = - \frac{Δ ϕ}{Δ t}$

$E \times 2 π R = - \frac{(0 - B \times π a^{2})}{Δ t}$

$E = \frac{B a^{2}}{2 R (Δ t)} .$

Tangential force on the ring,

$F = E q = \frac{B a^{2}}{2 R (Δ t)} \times (λ \times 2 π R)$

Torque, $τ = F R = \frac{B a^{2}}{2 R (Δ t)} \times (λ \times 2 π R) \times R$

Now, $τ = \frac{Δ L}{Δ t}$

$\frac{B a^{2} \times (2 π λ R)}{2 Δ t} = \frac{I ω}{Δ t} = \frac{(M R^{2}) ω}{Δ t}$

$∴ ω = \frac{B π a^{2} λ}{M R}$

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