A coil having 10 text{ A m}^2 magnetic moment | vedclass

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(C) The potential energy of a magnetic dipole in a magnetic field is given by $U = -\vec{m} \cdot \vec{B} = -mB \cos 	heta$.Initially, the axis of th

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$20$

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(C) The potential energy of a magnetic dipole in a magnetic field is given by $U = -\vec{m} \cdot \vec{B} = -mB \cos 	heta$.Initially, the axis of the coil is in the direction of the field, so $	heta_i = 0^{\circ}$.$U_i = -mB \cos 0^{\circ} = -mB$.After rotating through $90^{\circ}$, the final angle is $	heta_f = 90^{\circ}$.$U_f = -mB \cos 90^{\circ} = 0$.The change in potential energy is $\Delta U = U_f - U_i = 0 - (-mB) = mB$.This change in potential energy is converted into rotational kinetic energy: $\Delta U = K_f - K_i$.Since the coil starts from rest, $K_i = 0$, so $mB = \frac{1}{2} I \omega^2$.Rearranging for $\omega$: $\omega = \sqrt{\frac{2mB}{I}}$.Substituting the given values: $m = 10 	ext{ A m}^2$, $B = 2 	ext{ T}$, $I = 0.1 	ext{ kg m}^2$.$\omega = \sqrt{\frac{2 	imes 10 	imes 2}{0.1}} = \sqrt{\frac{40}{0.1}} = \sqrt{400} = 20 	ext{ rad/s}$.

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