Question 109: Magnetism and Matter

Question

The magnitude of magnetic induction at mid point $\mathrm{O}$ due to current arrangement as shown in Fig will be
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Accepted Answer

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}ight) = \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}ight) = \frac{\mu_0 I}{2 \pi a}
\end{aligned}
$
So, $B_{	ext {net }}=B_1+B_2=\frac{\mu_0 I}{\pi a}$

Answer

$\frac{\mu_{0} I}{\pi a}$

Answer

$\frac{\mu_{0} I}{4 \pi a}$

Answer

$\frac{\mu_{0} I}{2 \pi a}$

Answer

0

Magnetism and Matter

The magnitude of magnetic induction at mid point \mathrm{O} due to current arrangement as shown in Fig will be

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