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A current carrying loop is placed in a uniform magnetic field 'B' in different orientations I, II, III and IV as shown in the figure. The correct order of decreasing potential energy is
( $\hat{\mathrm{n}}$-unit vector normal to the plane of the loop)
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( $\hat{\mathrm{n}}$-unit vector normal to the plane of the loop)
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The correct answer is:
I, IV, II, III
Potential energy in a uniform magnetic field, $\mathrm{u}=-\mathrm{mB} \cos \theta$
(i) For orientation I $\theta=180^{\circ} \therefore \mathrm{u}=-\mathrm{mB} \cos 180^{\circ}=$ $\mathrm{mB}$
(ii) For orientation II $\theta=90^{\circ} \therefore \mathrm{u}=-\mathrm{mB} \cos 90^{\circ}=0$
(iii) For orientation III $\theta=0^{\circ} < \theta < 90^{\circ} \therefore-\mathrm{mB} < \mathrm{u} < 0$
(iv) For orientation IV $\theta=90^{\circ} < \theta < 180^{\circ} \therefore 0 < \mathrm{u} < $ $\mathrm{mB}$
(i) For orientation I $\theta=180^{\circ} \therefore \mathrm{u}=-\mathrm{mB} \cos 180^{\circ}=$ $\mathrm{mB}$
(ii) For orientation II $\theta=90^{\circ} \therefore \mathrm{u}=-\mathrm{mB} \cos 90^{\circ}=0$
(iii) For orientation III $\theta=0^{\circ} < \theta < 90^{\circ} \therefore-\mathrm{mB} < \mathrm{u} < 0$
(iv) For orientation IV $\theta=90^{\circ} < \theta < 180^{\circ} \therefore 0 < \mathrm{u} < $ $\mathrm{mB}$
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