A square coil of edge l having a turns carries a current i. It is kept on a smooth horizontal plate. A uniform magnetic field B exists in a direction parallel to an edge. The total mass of the coil is M. What should be the minimum value of B for which the coil will start tipping over.

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$\frac{m g}{2 n i l}$

$\frac{mg}{nil}$

$\frac{2 mg}{nil}$

$\frac{nil}{2 mg}$

answer is C.

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Detailed Solution

Torque produced by field must be able to balance the torque produced due to weight
$\begin{array}{l} τ = τ_{weight} \\ {Bil}^{2} n = mg \frac{l}{2} \\ \Rightarrow B = \frac{mgl}{2 {nil}^{2}} \Rightarrow B = \frac{mg}{2 nil} \end{array}$

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