A thin flexible wire of length is connected to two adjacent fixed points and carries a current in the clockwise direction, as shown in the figure. When the system is put in a uniform magnetic field of strength going into the plane of the paper, the wir

Question

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 [2010]

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(3)

Let us consider an elemental length $I$ subtending an angle $d θ$ at the centre of the circle $' O'$ . Let $F_{B}$ be the magnetic force acting on this length.

$∴$ $F_{B} = B I (d l) (upwards)$

$= B I (R d θ)$ $[∵ angle d θ = \frac{arc (d l)}{radius R}]$

$= B I (\frac{L}{2 π}) d θ$ $[∵ 2 π R = L \Rightarrow R = \frac{L}{2 π}]$

Let $T$ be the tension in the wire acting along both ends of the elemental length.

At equilibrium, $2 T \sin (\frac{d θ}{2}) = B I \frac{L}{2 π} d θ$

$\Rightarrow 2 T (\frac{d θ}{2}) = B I \frac{L}{2 π} d θ$ $[∵ \frac{d θ}{2} is very, very small]$

$∴$ $T = \frac{B I L}{2 π}$

Solution

1 $I B L$

2 $\frac{I B L}{π}$

3 $\frac{I B L}{2 π}$

4 $\frac{I B L}{4 π}$

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Solution

1 IBL

2 IBLπ

3 IBL2π

4 IBL4π

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1 $I B L$

2 $\frac{I B L}{π}$

3 $\frac{I B L}{2 π}$

4 $\frac{I B L}{4 π}$