Question

A wire carrying current I is tied between points P and Q and is in the shape of a circular arc 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)

$IBR$

Consider small element of length $d\overrightarrow{l}$subtending small angle $d\theta$ at origin of wire:

The magnetic force on the element is $F=Id\overrightarrow{l}\times \overrightarrow{B}=IdlB\hat{r}$ radially outward direction

The resultant of tension must balance magnetic force as wire element is in equilibrium

The component of tension in downward direction $T_{resultant}=2Tsin\frac{d\theta }{2}$

$T_{resultant}=F\Rightarrow 2Tsin\frac{d\theta }{2}=IdlB$

for small angle $\frac{d\theta }{2}$ the value of $sin\frac{d\theta }{2}$ can be approximated as $sin\frac{d\theta }{2}=\frac{d\theta }{2}$ and $d\theta =\frac{dl}{R}$ where R is radius of the arc.

$2T\frac{d\theta }{2}=IdlB\Rightarrow T\frac{dl}{R}=IdlB$

$T=IBR$.