An equilateral triangle loop ACD of side $l$ carries a current I in a direction shown in figure The loop is kept in a uniform horizontal magnetic field $\vec{B}$ as shown in figure. Choose the correct statements.

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Net force on the loop is zero

The magnitude of force on side AC, is $\frac{\sqrt{3}}{2} I l B$

The magnitude of force on side DC is zero

Net force on the loop is $\sqrt{3} I l B$

answer is A, C, D.

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Detailed Solution

$| F_{AC} | = I l B \sin 120^{0} = \frac{\sqrt{3}}{2} I l B (downward)$
$| F_{AD} | = I l B \sin 120^{0} = \frac{\sqrt{3}}{2} I l B (upward)$
$| F_{DC} | = 0$

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