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 wore. If the wire subtends a angle $2{\mathrm{\theta }}_0$ at the centre of the circle (of which it forms an arch) then the tension in the wire is :

Given data: 

current I, radius of the arch is R , magnetic field is B and the subtended angle is $2{\theta }_0$

Concept solution: Moving charges and magnetism

Detailed used:

Think about a little element length $\overrightarrow{\mathrm{dl}}$and a small angle d at the origin of the wire:

The magnetic force acting on the element is $\mathrm{F}=\mathrm{I}\overrightarrow{\mathrm{dl}}\times \overrightarrow{\mathrm{B}}=\mathrm{IdlB}\hat{\mathrm{r}},\mathrm{which} \mathrm{is} \mathrm{directed} \mathrm{radially} \mathrm{outward}.$

Magnetic force must be balanced as a result of tension since the wire is in equilibrium.

The down-directed tension component ${\mathrm{T}}_{\text{resultant }}=2 \mathrm{Tsin}\frac{\mathrm{d}\theta }{2}$

${\mathrm{T}}_{\text{resultant }}=\mathrm{F}\Rightarrow 2 \mathrm{Tsin}\frac{\mathrm{d}\theta }{2}=\mathrm{IdlB}$

When the angle $\frac{\mathrm{d}\theta }{2}$ is small, the value of $\sin \frac{\mathrm{d}\theta }{2}$can be roughly calculated $\sin \frac{d\theta }{2}=\frac{d\theta }{2}$

$\mathrm{and} \mathrm{d\theta }=\frac{\mathrm{dl}}{\mathrm{R}},\mathrm{where} \mathrm{R} \mathrm{is} \mathrm{the} \mathrm{radius} \mathrm{of} \mathrm{the} \mathrm{arc}.$

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

$\mathrm{T}=\mathrm{IBR}.$