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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 the figure. When the system is put in a uniform magnetic field of strength B going into the plane of the paper, the wire takes the shape of a circle. The tension in the wire is

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$\frac{IB L}{π}$

$\frac{IB L}{2 π}$

IBL

$\frac{IB L}{4 π}$

answer is C.

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Detailed Solution

Given data

Length= $l$

Current= $I$

Magnetic field=B

Concepts used moving charges and magnetism

Explanation

$2 T \sin \frac{d θ}{2} = B I R d θ$

$T d θ = B I R d θ$

$T = B I R$

$T = \frac{B I L}{2 π}$

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