Question

A wire carrying current $I$ is tied between points $P$ and $Q$ and is in the shape of a circular arch of radius $R$ due to a uniform magnetic field $B$ (perpendicular to the plane of the paper, shown by xxx) in the vicinity of the wire. If the wire subtends an angle $2\,\theta_{0}$ at the centre of the circle (of which it forms an arch) then the tension in the wire is :

• A. $IBR$
• B. $\frac{IBR}{\sin\theta_{0}}$
• C. $\frac{IBR}{2 \sin\theta_{0}}$
• D. $\frac{IBR\theta_{0}}{\sin\theta_{0}}$

Answer 1

Answer: (A)

For equilibrium
$2 T \sin \left(\frac{d \theta}{2}\right)=Bi\, dl$
$\Rightarrow 2 T\left(\frac{d \theta}{2}\right)= Bi(R d \theta)$
$\Rightarrow T=Bi R=$ constant