Question

In the given figure net magnetic field at $O$ will be

• A. $\frac{2 \mu_{0} i}{3 \pi a} \sqrt{4-\pi^{2}}$
• B. $\frac{\mu_{0} i}{3 \pi a} \sqrt{4+\pi^{2}}$
• C. $\frac{2 \mu_{0} i}{3 \pi a^{2}} \sqrt{4+\pi^{2}}$
• D. $\frac{2 \mu_{0} i}{3 \pi a} \sqrt{\left(4-\pi^{2}\right)}$

Answer 1

Answer: (B)

Magnetic field at 0 due to Part (1) : $B_{1}=0$

Part (2): $B_{2}=\frac{\mu_{0}}{4 \pi} \cdot \frac{\pi i}{(a / 2)} \otimes$ (along $-Z$-axis)
Part (3): $B_{3}=\frac{\mu_{0}}{4 \pi} \cdot \frac{i}{(a / 2)}(\downarrow)$ (along $-Y$-axis)
Part (4): $B_{4}=\frac{\mu_{0}}{4 \pi} \cdot \frac{\pi i}{(3 a / 2)} \odot$ (along $+Z$-axis)
Part (5): $B_{5}=\frac{\mu_{0}}{4 \pi} \cdot \frac{i}{(3 a / 2)}(\downarrow)$ (along $-Y$-axis)
$B_{2}-B_{4}=\frac{\mu_{0}}{4 \pi} \cdot \frac{\pi i}{a}\left(2-\frac{2}{3}\right)=\frac{\mu_{0} i}{3 a} \otimes$ (along $-Z$-axis)
$B_{3}+B_{5}=\frac{\mu_{0}}{4 \pi} \cdot \frac{1}{a}\left(2+\frac{2}{3}\right)=\frac{8 \mu_{0} i}{12 \pi a}(\downarrow)$ (along $-Y$-axis)
Hence net magnetic field
$B_{\text {net }}=\sqrt{\left(B_{2}-B_{4}\right)^{2}+\left(B_{3}+B_{5}\right)^{2}}=\frac{\mu_{0} i}{3 \pi a} \sqrt{\pi^{2}+4}$