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A square coil of side ‘a’ carrying current $' i'$ and is having one of its side AB parallel to y-axis and its plane is at angle $θ = 30 °$ with x-axis (as shown). If a uniform magnetic field B exist in the region along $\hat{k}$ direction, then torque due to magnetic force on the coil is:

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\frac{i a^{2} B}{2} \hat{j}

\frac{i a^{2} B}{\sqrt{2}} (- \hat{k} + \hat{j})

- \frac{i a^{2} B}{2} \hat{j}

\frac{i a^{2} B}{2} \hat{i}

answer is A.

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Detailed Solution

\vec{M} = i a \hat{a} = i a^{2} (\cos 30 ° \hat{k} - \sin 30 ° \hat{i})

$M = \frac{i a^{2}}{2} (\sqrt{3} \hat{k} - \hat{i})$

$\bar{B} = B \hat{k}$

$\vec{τ} = \vec{M} \times \vec{B}$

$= \frac{i a^{2} B}{2} \hat{j}$

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