Question

The lower end of a rigid rectangular coil $($sides a and b$)$ is hanging in a uniform unknown magnetic field $B$ (into the plane of paper) & the upper end is hung from the arm of $a$ balance. A current $i$ is set up in the coil in the direction shown and weights are placed in the right hand balance pan until the system is balanced. Now the direction of current is reversed keeping magnitude same and a mass $m$ had to be added to the left balance pan for balance. Magnetic field $B$ is :

• A. $\frac{mg}{ia}$
• B. $\frac{mg}{2ia}$
• C. $\frac{2mg}{ia}$
• D. $\frac{mg}{4ia}$

Answer 1

Answer: (B)

$\frac{mg}{2ia}$

Force on arm of length a
$F_a=iaB$ in the direction opposite to the weight $\left(mg\right)$
Since current in two arms having length b are opposite to each other, force on one side will cancel the force on other side.
When the current direction is changed, a force $F_a=iaB$ will act in the same direction.
Thus net change in force is $F_{net}=2iaB=mg$ when mg is the weight added.
$\therefore B=\frac{mg}{2ia}$