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Magnetism

Section: Physics

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Question : 41 of 46

Marks: +1, -0

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

[JEE Adv 2010 P1]

IBL
$\frac{IBL}{\pi}$
$\frac{IBL}{2\pi}$
$\frac{IBL}{4\pi}$

Solution:

From the figures depicted here, for equilibrium of the small element, we have dl = Rdθ. Therefore,

2T sin

\left(\frac{d\theta}{2}\right)

= BI (Rdθ)

Since dθ << sin

\left(\frac{d\theta}{2}\right)

=

\frac{d\theta}{2}

, we have

T dθ = BIR dθ ⇒ T = BIR

Also, it is given that L = 2πR.

Hence,

R =

\frac{L}{2\pi}

Therefore,

T =

\frac{BIL}{2\pi}

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