Q148 / 160 · TS EAMCET 2023 (14 May Shift 2) · Physics

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) Figure

  • A
    I, III, II, IV
  • B
    I, II, III, IV
  • Correct answer
    C
    I, IV, II, III
  • D
    III, IV, I, II

Solution

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} < $ $\mathrm{mB}$

Hence, the correct answer is option: 'C'.

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