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. When the system is put in uniform magnetic field of strength $B$ going into the plane of the paper, the wire take the shape of a circle. The tension in the wire, after acquiring the circular shape is

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Aicharya · Accepted Answer

Given that, the length of wire $=L$ Current passing through ring $=I$ Let $R$ be the radius of ring. Then, $2 \pi R=L$ $ R=\frac{L}{2 \pi}...(i) $ Consider an element of ring $A B$ which subtends an angle $\Delta \theta$ at the centre $O$, such that, $A B=\Delta l=R \Delta \theta$...(ii) If $T$ be the tension at point $P$ as shown in figure.

We know that, force acting on element $A B$ in uniform magnetic field, $ F=I B \Delta l $ ... (iii) (radially outward) By resolving tension into sine and cosine components $T \cos \frac{\Delta \theta}{2}$ is acting equal and opposite. So, it is cancelled.

But sum of sine component is equal and opposite to $F$. $\therefore$ At equilibrium, we get $ 2 T \sin \frac{\Delta \theta}{2}=F $ As $\frac{\Delta \theta}{2}$ is very small angle, so, $\sin \frac{\Delta \theta}{2}=\frac{\Delta \theta}{2}$ By substituting the values, we get $ \begin{aligned} 2 T \frac{\Delta \theta}{2} & =I B \Delta l \Rightarrow T \Delta \theta=I B R \Delta \theta \\ T & =I B R...(i) \end{aligned} $ Using Eqs. (i) and (iv), we get $ T=I B L / 2 \pi $

A thin flexible wire of length $L$ is connected to two adjacent fixed points and carries a current $I$ in…

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