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