Question

A thin uniform rod with negligible mass and length $l$ is attached to the floor by a frictionless hinge at its one point P as shown in the figure. I is the current in the rod A horizontal spring of spring constant k connects a vertical wall and other end Q of the rod. The whole system is kept in a horizontal uniform magnetic field whose magnitude is B (gravity is absent in the space).

The torque due to magnetic force on the rod about point P is :

• A. $IBl$
• B. $2IB{l}^2$
• C. $\frac{IB{l}^2}{2}$
• D. Zero

Answer 1

The correct option is C $\frac{IB{l}^2}{2}$

As shown in the above fig,

The magnetic force on the section $dr$ is,

$d{F}_1=IBdr$, magnetic field $B$ and section length $dr$ are perpendicular to each other.

Force is also perpendicular to the rod.

The torque due to magnetic force on this section is,

$\tau =rd{F}_1=IBrdr$

The total torque is,

$\int d\tau =IB{\int }_0^{l}rdr$

$\tau =\frac{I{l}^{2}B}{2}$.