Question

A thin uniform rod with negligible mass and length $0.2\text{m}$ is attached to the floor by a frictionless ring at point $\text{P}$. A horizontal spring with spring constant $k=4.80{\text{Nm}}^{-1}$ connected to rod, the other end of the spring is fixed with a vertical wall as shown in the figure below. The rod is in a uniform magnetic field $B=0.340\text{T}$ directed into the plane of figure. There is a current $I=6.50\text{A}$ in the rod in the direction shown. Find the energy stored in the spring when the rod is in equilibrium:

• A. $4.8\times {10}^{-3}\text{J}$
• B. $3.5\times {10}^{-3}\text{J}$
• C. $7.8\times {10}^{-3}\text{J}$
• D. $6.2\times {10}^{-3}\text{J}$

Answer 1

The correct option is C $7.8\times {10}^{-3}\text{J}$

The magnetic force acting on rod will be,

$\overrightarrow{F}=i(\overrightarrow{L}\times \overrightarrow{B})$

$F=BiL\sin {90}^{\circ }=BiL$


Considering the magnetic force to be acting at (C.O.M) of the uniform rod as shown in figure, its torque about the hinge point is,

$\tau =\left(BiL\right)\left(\frac{L}{2}\right)=\frac{BI{L}^2}{2}$

$\tau =\frac{\left(0.340\right)\left(6.50\right)(0.2{)}^2}{2}=0.0442\text{Nm}$

For equilibrium of the rod,

Clockwise torque of magnetic force = Anticlockwise torque of spring force

$\Rightarrow 0.0442=\left(kx\right)L\sin {53}^{\circ }$

$0.0442=x\left(4.80\right)\times \left(0.2\right)\left(\frac{4}{5}\right)$

$x=\frac{0.0442\times 5}{4.8\times 0.2\times 4}=0.057\text{m}$

The energy stored in spring will be,

$U=\frac{1}{2}k{x}^2$

$U=\frac{1}{2}\times 4.80\times (0.057{)}^2$

$U=7.8\times {10}^{-3}\text{J}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence,option (c) is the correct answer.