A thin flexible wire of length L is connected | vedclass

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(C) Given that,the length of the wire is $L$. The current passing through the wire is $I$. Let $R$ be the radius of the circular loop formed by the wi

Vedclass · Accepted Answer

$\frac{I B L}{2 \pi}$

Vedclass · Answer

(C) Given that,the length of the wire is $L$. The current passing through the wire is $I$. Let $R$ be the radius of the circular loop formed by the wire.Then,the circumference $2 \pi R = L$,which gives $R = \frac{L}{2 \pi}$.Consider a small element of the wire $AB$ of length $\Delta l$ which subtends an angle $\Delta 	heta$ at the center $O$. Thus,$\Delta l = R \Delta 	heta$.The magnetic force acting on this small element $AB$ is $F = I B \Delta l = I B R \Delta 	heta$,directed radially outward.Let $T$ be the tension in the wire. The components of tension $T \cos(\Delta 	heta / 2)$ at the ends of the element cancel each other out.The radial component of the tension force is $2 T \sin(\Delta 	heta / 2)$.For equilibrium,the magnetic force must be balanced by the radial component of the tension:$2 T \sin(\Delta 	heta / 2) = F = I B R \Delta 	heta$.Since $\Delta 	heta$ is very small,$\sin(\Delta 	heta / 2) \approx \Delta 	heta / 2$.Therefore,$2 T (\Delta 	heta / 2) = I B R \Delta 	heta$,which simplifies to $T = I B R$.Substituting $R = \frac{L}{2 \pi}$ into the expression for tension,we get $T = I B (\frac{L}{2 \pi}) = \frac{I B L}{2 \pi}$.

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