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Standard XII

Physics

Magnetic Field Due to a Circular Arc at the Centre

A thin flexib...

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:

I B L

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\frac{I B L}{π}

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\frac{I B L}{2 π}

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\frac{I B L}{4 π}

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Solution

The correct option is B $\frac{I B L}{2 π}$
$2 T sin \frac{d θ}{2} = B i R d θ$
$T d θ = B i R d θ$ (for $θ$ small)
$T = B i R = \frac{B i L}{2 π}$

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The correct option is B IBL2π2Tsindθ2=BiRdθTdθ=BiRdθ (for θ small)T=BiR=BiL2π

The correct option is B $\frac{I B L}{2 π}$
$2 T sin \frac{d θ}{2} = B i R d θ$
$T d θ = B i R d θ$ (for $θ$ small)
$T = B i R = \frac{B i L}{2 π}$