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 2θ₀ at the centre of the circle (of which it forms an arch) then the tension in the wire is :

Consider small element of length d→l subtending small angle dθ at origin of wire:
The magnetic force on the element is
F = Id→l × →B = IdlB^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
Tresultant = 2Tsin(dθ/2)
2Tresultant = F → 2Tsin(dθ/2) = IdlB
for small angle dθ/2 the value of sin(dθ/2) can be approximated as sin(dθ/2) = dθ/2 and dθ = dl/R where R is radius of the arc.
2T(dl/2R) = IdlB → Tdl/R = IdlB
T = IBR.