JEE MAIN - Physics (2023 - 29th January Morning Shift - No. 6)
The magnitude of magnetic induction at mid point $$\mathrm{O}$$ due to current arrangement as shown in Fig will be
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$$\frac{\mu_{0} I}{\pi a}$$
$$\frac{\mu_{0} I}{4 \pi a}$$
$$\frac{\mu_{0} I}{2 \pi a}$$
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Explanation
Magnetic field due to wire $A B$ and wire $E D$ cancel each other. So, magnetic field due to BC and ET will be
$$ \begin{aligned} & B_1=\frac{\mu_0 I}{4 \pi(a / 2)} \cdot\left(\sin \frac{\pi}{2}+\sin 0^{\circ}\right) = \frac{\mu_0 I}{2 \pi a} \\ & B_2=\frac{\mu_0 I}{4 \pi(a / 2)} \cdot\left(\sin \frac{\pi}{2}+\sin 0^{\circ}\right) = \frac{\mu_0 I}{2 \pi a} \end{aligned} $$
So, $B_{\text {net }}=B_1+B_2=\frac{\mu_0 I}{\pi a}$
$$ \begin{aligned} & B_1=\frac{\mu_0 I}{4 \pi(a / 2)} \cdot\left(\sin \frac{\pi}{2}+\sin 0^{\circ}\right) = \frac{\mu_0 I}{2 \pi a} \\ & B_2=\frac{\mu_0 I}{4 \pi(a / 2)} \cdot\left(\sin \frac{\pi}{2}+\sin 0^{\circ}\right) = \frac{\mu_0 I}{2 \pi a} \end{aligned} $$
So, $B_{\text {net }}=B_1+B_2=\frac{\mu_0 I}{\pi a}$
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