Figure shows square current carrying coil of edge length L. The magnetic field on the coil is given by $\vec{B} = \frac{B_{0} y}{L} \hat{i} + \frac{B_{0} x}{L} \hat{j}$ where B₀ is a positive constant.

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coil is free to rotate about x axis torque on the coil is given by $\frac{1}{2} {iAB}_{0} \hat{i}$

If coil is free to rotate about y-axis torque on coil is given by $- \frac{1}{2} {iAB}_{0} \hat{j}$

Resultant force on coil is zero.

Equation for the torque $\vec{μ} \times \vec{B}$ where $μ$ is magnetic moment of coil, is not valid here.

answer is A, B, C, D.

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Detailed Solution

Magnitude of force acting on each of the sides parallel to the x axis = $\frac{1}{2} i B_{0} L . T h e s e f o r c e s a r e o p p o s i t e l y d i r e c t e d a n d p e r p e n d i c u l a r t o t h e p l a n e o f t h e l o o p .$ So their resultant is zero.Force on the side lying along x-axis is directed into the plane of the loop and the force acting on the other side which is parallel to x - axis is directed out of the plane of the loop. These two forces form a couple which is directed along positive x-axis .

Magnitude of force acting on each of the sides parallel to y axis is $\frac{1}{2} i B_{0} L$ .These forces are antiparallel and perpendicular to the plane of the loop .Force on the side lying on the y axis is directed into the page and the force acting on the other side which is parallel to the y axis is directed out of the plane of the page .So the couplle acting on the loop is directed along negative y axis and has a magnitude $(\frac{1}{2} i B_{0} L) \times L i . e . \frac{1}{2} i B_{0} A .$