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.


When the rod is in equilibrium and makes an angle of ${53}^{\circ }$ with the floor then spring length changes by -

• A. $0.05765\text{m}$, stretched
• B. $0.05765\text{m}$, compressed
• C. $0.02421\text{m}$, stretched
• D. $0.02421\text{m}$, compressed

Answer 1

The correct option is A $0.05765\text{m}$, stretched

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,

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

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

As the torque developed by magnetic force is clockwise, the spring is stretched.

This torque will be balanced by the torque produced by the spring force.


Let the extension in spring be $x$.

So,

$\left(kx\right)L\sin \alpha =\frac{IB{L}^2}{2}$

$\Rightarrow 4.80\times x\times 0.200\times \sin {53}^{\circ }=\frac{6.5\times 0.340\times {0.200}^2}{2}$

$\Rightarrow x=0.05765\text{m}$ $($stretched$)$