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Q. A thin flexible wire of length $L$ is connected to a battery (negligible in size as compared to the wire) to form a closed loop. The battery drives a current $I$ in the wire. When the system is put in a uniform magnetic field of strength $B$ perpendicular to the plane in which the wire is placed, the wire takes the shape of a circle. The tension in the wire is

NTA AbhyasNTA Abhyas 2022

Solution:

$L=2\pi R$
$ \, R=\frac{L}{2 \pi }$
$2Tsin \left(d \theta \right) = F_{m}$

Solution
For small angles, $sin \left(\right. d \theta \left.\right) \approx d \theta $
$ \, 2T\left(d \theta \right)=I\left(d L\right)Bsin 90 ^\circ $
$ \, =I \, \left(R \, . d \theta \right).B$
$ \, T=IRB=\frac{I L B}{2 \pi }$