Question

A thin, uniform rod with negligible mass and length $0.200m$ is attached to the floor by a frictionless hinge at point $P($as shown in fig$)$. A horizontal spring with force constant $k=4.80N{m}^{-1}$ connects the other end of the rod to a vertical wall. The rod is in a uniform magnetic field $B=0.340T$ directed into the plane of the figure. There is current $I=6.50A$ in the rod, in the direction shown. When the rod is in equilibrium and makes an angle of ${53.0}^o$ with the floor, is the spring stretched or compressed?

• A. $0.05765m,$ stretched
• B. $0.05765m,$ compressed
• C. $0.0242m,$ stretched
• D. $0.0242m,$ compressed

Answer 1

The correct option is A $0.05765m,$ stretched

At equilibrium condition (as shown picture)
As current is flowing in $x-y$ plane and $\overrightarrow{B}$ is along z direction so angle between I$\overrightarrow{l}$ & $\overrightarrow{B}$ is $90°$
Force $F_B$ is perpendicular to radius as shown
$\overrightarrow{F_B}=I(\overrightarrow{l}\times \overrightarrow{B})\Rightarrow F_B=ILB\sin 90°=ILB$
By using Fleming's left hand rule we can find direction of $F_B$ and is already shown in figure
Due to $F_B$ the rod rotates in clockwise direction
So the spring gets stretched
So a force $kx$ is acting as shown
At equilibrium conditions moment at point P due to $F_B$= moment at point P due to spring force
$\Rightarrow \frac{\overrightarrow{l}}{2}\times \overrightarrow{F_B}=\overrightarrow{l}\times \left(k\overrightarrow{x}\right)$
$\Rightarrow \frac{l}{2}{F}_B\sin 90=lkx\sin (180-53°)$
$\Rightarrow \frac{F_B}{2}=kx\sin 53$
$\Rightarrow \frac{ILB}{2}=kx\sin 53$
$\Rightarrow x=\frac{ILB}{2k\sin 53}$
$\Rightarrow x=\frac{6.5\times 0.2\times 0.34}{2\times 4.8\sin 53}$
$=\frac{0.04604}{0.7986}$
$x=0.05764m$ stretched
Answer A