Question

A square coil of edge $l$ having $n$ turns carries a current $i$. It is placed on a smooth horizontal plate A magnetic field $B$ parallel to one edge is applied. The total mass of coil is $M$. The minimum value of $B$ for which the coil will tip over is

• A. $\frac{Mg}{lin}$
• B. $\frac{Mg}{2lin}$
• C. $\frac{2Mg}{lin}$
• D. none of these

Answer 1

The correct option is B $\frac{Mg}{2lin}$

Let the magnetic field be applied along AB.
Forces on side AB and CD are zero respectively.
Force on each side BC and DA is $F=nilB$ where n is the no of turns and B is the magnetic field and l is the length of each side of the square coil.
Total force is $F_{total}=2nilB$
$F_{total}$ balances the weight Mg of the coil.
$\therefore 2nilB=Mg$
$B=\frac{Mg}{2nil}$