Q. 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
A) $IBL$
B) $\frac{IBL}{\pi}$
C) $\frac{IBL}{2\pi}$
D) $\frac{IBL}{4\pi}$
Explanation
(c) $L = 2\pi R$
$\therefore\ \ R = \frac{L}{2\pi}$
$2T\sin{(d\theta) = F_{m}}$
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For small angles, $\sin{(d\theta) \approx d\theta}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ 2T(d\theta) = I\ (dL)B\sin{90{^\circ}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = I\ (2R\ .d\theta).B$
⦁ $\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T = IRB = \frac{ILB}{2\pi}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.