Q. A long wire bent as shown in the figure carries current $I=10A$ . If the radius of the semicircular portion is $1m$ , the magnetic induction (in $μT$ ) at the centre $C$ is

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A

$\text{\pi }^{2} + 4$

B

$\sqrt{\text{\pi }^{2} + 4}$

C

$\text{\pi }^{2} - 4$

D

$\sqrt{\text{\pi }^{2} - 4}$

Solution:

B₁ (due to two semi-infinite straight wires) $=\frac{\left(\text{μ}\right)_{0}}{2 \text{\pi }}\left(\frac{\text{I}}{a}\right)\hat{\text{k}}=2\hat{k}μT$
B₂ (semi-circular loop) $=\frac{\left(\text{μ}\right)_{0} \text{I}}{4 a}\left(- \hat{\text{i}}\right)=-\pi \hat{i}μT$
$\left|\overset{ \rightarrow }{\text{B}}\right| = \sqrt{\text{B}_{1}^{2} + \text{B}_{2}^{2}}$
$\left|\overset{ \rightarrow }{B}\right|=\sqrt{\pi ^{2} + 4}μT$

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