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In the given figure, what is the magnetic field induction at point $O$.
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Verified Answer
The correct answer is:
$\frac{\mu_0 I}{4 r}+\frac{\mu_0 I}{4 \pi r}$
Magnetic field due to straight wire above $O$ is zero i.e. $B_1=0$ (since, $\theta=0^{\circ}$ )
The magnetic field due to semi circular part
$B_2=\left(\frac{\mu_0 n I}{2 r}\right)=\frac{\mu_0(1 / 2) I}{2 r}=\frac{\mu_0 I}{4 r}$
The magnetic field due to lower straight portion
$B_3=\frac{1}{2} \frac{\mu_0 I}{2 \pi r} \text { (upward) }$
Net magnetic field $B=B_1+B_2+B_3$
$=0+\frac{\mu_0 I}{4 r}+\frac{\mu_0 I}{4 \pi r} \text { (upwards) }$
The magnetic field due to semi circular part
$B_2=\left(\frac{\mu_0 n I}{2 r}\right)=\frac{\mu_0(1 / 2) I}{2 r}=\frac{\mu_0 I}{4 r}$
The magnetic field due to lower straight portion
$B_3=\frac{1}{2} \frac{\mu_0 I}{2 \pi r} \text { (upward) }$
Net magnetic field $B=B_1+B_2+B_3$
$=0+\frac{\mu_0 I}{4 r}+\frac{\mu_0 I}{4 \pi r} \text { (upwards) }$
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