Question

A thin flexible wire of length $L$ is connected to two close adjacent fixed points and carries a current $I$ in the clockwise direction, as shown in the figure. When the system is put in a uniform magnetic field of strength $B$ going into the plane of the paper, the wire takes the shape of a circle. The tension in the wire is

• A. $IBL$
• B. $\frac{IBL}{\pi }$
• C. $\frac{IBL}{2\pi }$
• D. $\frac{IBL}{4\pi }$

Answer 1

The correct option is C $\frac{IBL}{2\pi }$

Free body digram:


If we take any small segment of the loop,
Force due to magnetic field $d{F}_B=I(\overrightarrow{dl}\times \overrightarrow{B})$

$F_B=I\left[\int dl\right]B\sin {90}^{\circ }=IBL$ (direction shown in diagram)

From figure we can see that

$2T\sin \theta =I\left(dl\right)B$ (as all the forces are balanced)

For small values of $\theta$ : $\sin \theta \simeq \theta$

$\Rightarrow 2T\theta =IdlB$

$\Rightarrow 2T\theta =I\left(2\theta R\right)B[\because dl=R(2\theta \left)\right]$

$\Rightarrow T=IRB$

As, $L_{\text{arc length}}=2\pi R$ = Circumference

$\therefore R=\frac{L}{2\pi }$

$\Rightarrow T=\frac{IBL}{2\pi }$

Hence, option (c) is the correct answer.