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A conducting ring of mass 2 kg and radius 0.5 m is placed on a smooth horizontal plane. The ring carries a current I=4 A. A horizontal magnetic field $B = 10 T$ is switched on at time t=0 as shown in figure. The ring is left free to move. The angular acceleration of the ring at t =1,sec is $β \times π rad / s^{2}$ . The value of $β$ is

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answer is 40.

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Detailed Solution

$\begin{array}{l} \vec{M} = IA (- \hat{k}) = - (4) (π) (0 .5)^{2} \hat{k} \\ \Rightarrow \vec{M} = π \hat{k} ({Am}^{2}) \end{array}$

Since, $τ = \vec{M} \times \vec{B} = (- π \hat{k}) \times (10 \hat{i}) = - (10 π) \hat{j}$

The axis of rotation is along $\vec{τ}$ i.e., axis of rotation is the y-axis, moment of inertia
about which is

$I = \frac{1}{2} {mR}^{2} = \frac{1}{2} (2) (0 .5)^{2} = \frac{1}{4} {kgm}^{2} \Rightarrow α \frac{| \bar{τ} |}{I} = \frac{10 π}{(\frac{1}{4})} = 40 π {rads}^{- 2}$

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