Question

A uniform bar of length $L$ carries a current $I$ in the direction from point $\text{a}$ to $\text{b}$. The bar is in uniform magnetic field directed normally inwards. Torque acting on the rod due to magnetic field, about point $\text{a}$, is

• A. $BIL$
• B. $\frac{BIL}{2}$
• C. $BI{L}^2$
• D. $\frac{BI{L}^2}{2}$

Answer 1

The correct option is D $\frac{BI{L}^2}{2}$

Consider an infinitesimally small part of the length of the bar $dx$, as shown in the figure. The force acting on the length is $dF$ acting perpendicular to length of the rod. Torque of $dF$ about point 'a' is


$d\tau =dF.x$

Here, $dF=BIdx$

$\therefore d\tau =BIx.dx$

The total torque acting on the bar is,

$\tau =\int d\tau ={\int }_{x=0}^{x=L}BIx.dx$

$\tau =BI{\int }_{x=0}^{x=L}x.dx$

$\tau =BI{\left[\frac{x^2}{2}\right]}_0^L$

$\tau =\frac{BI{L}^2}{2}$

Hence, option $\left(D\right)$ is the correct answer.