Question

A loop of flexible conducting wire of length 0.5 m lies in a magnetic field of 1.0 T perpendicular to the plane of the loop. Show that when a current as shown in fig. is passed through the loop, it opens into a circle. Also calculate the tension developed in the wire if the current is 1.57 A

Answer 1

The loop may be divided into a large number of small length elements. When a current I is passed through the loop placed in the magnetic field such that the plane of the loop is perpendicular to field, then force on each element is

$d\overrightarrow{F}=Id\overrightarrow{l}\times \overrightarrow{B}=IdlBsin{90}^o$$=IdlB$

Perpendicular to current element $I\overrightarrow{dl}$ as well as magnetic field

$\overrightarrow{B}$. Hence, the loop opens into a circle.

Consider an element of length dl of circle of radius r, making an angle $alpha$ at centre.

If T is the tension in the wire, then force toward centre:

$2Tsin\frac{\alpha }{2}=IBdl$

For small angle $\alpha ,sin\frac{\alpha }{2}=\frac{\alpha }{2}$

$2T\cdot \frac{\alpha }{2}=IBdl$

$T=\frac{IBdl}{\alpha }=IBr(\therefore \alpha =\frac{dl}{r})$

$=IB\left(\frac{1}{2\pi }\right)(\therefore l=2\pi r)$

$=1.57\times 1\times \left(\frac{0.5}{2\times 3.14}\right)=0.125N$