A rectangular wire-loop of width $a$ is suspended from the insulated pan of a spring balance as shown in figure. A current $I$ exists in the anticlockwise direction in the loop. A magnetic field $B$ exists in the lower region. Find the change in the tension of the spring if the current in the loop is reversed.

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$F = \sqrt{3} B_{0} i l$

$F = \sqrt{2} B_{0} i l$

$F = \sqrt{2} B_{0} i l^{2}$

$F = \sqrt{2} B_{0} i^{2} l$

answer is B.

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Detailed Solution

$\vec{F} = \vec{i l} \times \vec{B} = i [l \hat{i} \times B_{0} (\hat{i} + \hat{j} + \hat{k})] = B_{0} i l (\hat{k} - \hat{j}) o r, F = \sqrt{2} B_{0} i l$

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