Question

A thin, uniform rod with negligible mass and length $0.200\text{m}$ is attached to the floor by a frictionless hinge at point $P$ $($as shown in the figure$)$. A horizontal spring with force constant $k=4.80{\text{Nm}}^{-1}$ connects the other end of the rod to a vertical wall. The rod is in a uniform magnetic field $B=0.340\text{T}$ directed into the plane of the figure. There is current $I=6.50\text{A}$ in the rod, in the direction shown.


Calculate the torque due to the magnetic force on the rod, for an axis at $P$.

• A. $0.0442\text{N-m}$, clockwise
• B. $0.0442\text{N-m}$, anticlockwise
• C. $0.0221\text{N-m}$, clockwise
• D. $0.0221\text{N-m}$, anticlockwise

Answer 1

The correct option is A $0.0442\text{N-m}$, clockwise

By examining a small piece of the wire of length $dr$, we find,


Magnetic force on the small piece,

$\overrightarrow{dF}=I\overrightarrow{dr}\times \overrightarrow{B}=IBdr\sin \theta \hat{\eta }$

$($direction as shown in the figure$)$

$\Rightarrow \overrightarrow{dF}=IBdr\sin {90}^{\circ }\hat{\eta }$$=IBdr\hat{\eta }$

Now, torque on the small piece,

$\overrightarrow{d\tau }=\overrightarrow{r}\times \overrightarrow{dF}=rdF\sin \theta$

$($clockwise$)$

$\Rightarrow d\tau =r\left(IBdr\right)\sin {90}^{\circ }$

$\Rightarrow d\tau =IBrdr$

So, torque on the rod,

$\tau ={\int }_0^{L}IBrdr$

$\Rightarrow \tau =\frac{IB{L}^2}{2}$

$\Rightarrow \tau =\frac{6.5\times 0.340\times {0.2}^2}{2}$

$\Rightarrow \tau =0.0442\text{N-m}$, clockwise