Question

A loop of flexible conducting wire of length $l$ lies in magnetic field B which is normal to the plane of loop. A current I is passed through the loop. The tension developed in the wire to open up is:

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

Answer 1

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

Let a small segment of length dl subtends an angle $d\theta$ at the center of the circle( radius R) of which dl is a small arc.
The tensions at both ends of dl will be equal in magnitude.
The horizontal components cancel each other.
The vertical components are downward.
Total vertical component: $2T\sin \left(\frac{d\theta }{2}\right)$
The upward force on the segment dl is IBdl
But, $dl=R\times \frac{d\theta }{2\pi }$
$\therefore IBR\times d\theta =2T\sin \left(\frac{d\theta }{2}\right)=2T\frac{d\theta }{2}$
$\Rightarrow T=IBR$ ....(2)
The shape of the wire will become a circle.
$\therefore l=2\pi R$
$\Rightarrow R=\frac{l}{2\pi }$
Substituting $R=\frac{l}{2\pi }$ in eqn(2)
$T=\frac{IBl}{2\pi }$