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): $B_4=\frac{{\mu }_0}{4\pi }\cdot \frac{\pi i}{(3a/2)}\odot$ (along $+Z$-axis)
Part (5): $B_5=\frac{{\mu }_0}{4\pi }\cdot \frac{i}{(3a/2)}(\downarrow )$ (along $-Y$-axis)
$B_2-B_4=\frac{{\mu }_0}{4\pi }\cdot \frac{\pi i}{a}\left(2-\frac{2}{3}\right)=\frac{{\mu }_{0}i}{3a}\otimes$ (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
$B_{\text{net }}=\sqrt{{\left(B_2-B_4\right)}^2+{\left(B_3+B_5\right)}^2}=\frac{{\mu }_{0}i}{3\pi a}\sqrt{{\pi }^2+4}$