Question

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

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

Answer 1

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

The moment wire of length $L$ is placed in magnetic field, magnetic forces acts in radially outward direction and the wire takes a circular shape.


Now, we know that for any small segment of circular wire subtending angle $d\theta$, the tension will act in symmetrical manner as shown below,


Balancing the forces we get,

$2T\sin \left(\frac{d\theta }{2}\right)=IdlB$

$\sin \frac{d\theta }{2}=\frac{d\theta }{2}$, for small angle

$2T\left(\frac{d\theta }{2}\right)=I\left(Rd\theta \right)B(\because dl=Rd\theta )$

$T=BIR$

For the loop in circular form,

$2\pi R=L$

or, $R=\frac{L}{2\pi }$

$\Rightarrow T=BI\left(\frac{L}{2\pi }\right)$

$\therefore T=\frac{BIL}{2\pi }$

Hence, option $\left(c\right)$ is the correct answer.