A uniform magnetic induction field $B_{0}$ exists in a cylindrical region of radius R. A conducting ring of radius a (>R) is placed co-axially with the field region. The ring has mass ‘m’ and linear charge density $λ$ . The angular velocity gained by the ring if the field is switched off is

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$\frac{λ π B_{0} a^{2}}{m R}$

$\frac{λ π B_{0} R^{2}}{2 m a}$

$\frac{λ π B_{0} R^{2}}{m a}$

$\frac{λ π B_{0} a^{2}}{2 m R}$

answer is C.

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Detailed Solution

$E = \frac{R^{2}}{2 a} \frac{d B}{d t}$
   $τ = λ 2 m a E a = λ 2 π a \frac{R^{2}}{2 a} \frac{d B}{d t} a$
   $τ = λ π R^{2} \frac{d B}{d t} a$
   $\int τ d t = λ π R^{2} a \int_{B_{0}}^{0} d B$
Lt $L_{f} - L_{i} = - λ π R^{2} B_{0} a$
   $I w = - λ π R^{2} B_{0} a$
   $w = \frac{λ π R^{2} B_{0}}{m a}$