Question

A square coil of edge $l$ having $n$ 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 $m$. The minimum value of $B$ for which the coil will start tipping over is $\frac{1}{k}$, what is the value of $k$ ? Given that $mg=nil$ (All physical quantities are in SI unit)

Answer 1

Answer: 2

Edge $=1$ , current $=$ i Turns $=n$ , mass $=m$ Magnetic field $=B$
$\tau =\mu Bsin90^{\circ }=\mu B$
Min Torque produced must be able to balance the torque produced due to weight Now, $\tau B$ weight $\mu B=mg\left(\frac{1}{2}\right)\Rightarrow n\times i\times l^{2}B=mg\left(\frac{1}{2}\right)$
$\Rightarrow B=\frac{mg}{2nil}=\frac{1}{2}$