JEE Advanced · Physics · 20. Magnetism & Current
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 \(I B L\)
- B \(\frac{I B L}{\pi}\)
- C \(\frac{I B L}{2 \pi}\)
- D \(\frac{I B L}{4 \pi}\)
Answer & Solution
Correct Answer
(C) \(\frac{I B L}{2 \pi}\)
Step-by-step Solution
Detailed explanation

\(L=2 \pi R\)
\(R=\frac{L}{2 \pi}\)
\(2 T \sin (d \theta)=F_m\)
for small angles, \(\sin (d \theta) \approx d \theta\)
\(\begin{aligned}
\therefore 2 T(d \theta) & =I(d L) B \sin 90^{\circ} \\
& =I(2 R \cdot d \theta) \cdot B \\
\therefore T & =I R B=\frac{I L B}{2 \pi}
\end{aligned}\)
\(\therefore\) correct option is (c)
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