Question

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

• A. $\frac{2{\mu }_{o}I}{3\pi a}\sqrt{4-{\pi }^2}$
• B. $\frac{{\mu }_{o}I}{3\pi a}\sqrt{4+{\pi }^2}$
• C. $\frac{4{\mu }_{o}I}{3\pi a}\sqrt{4-{\pi }^2}$
• D. $\frac{{\mu }_{o}I}{3\pi a}\sqrt{4-{\pi }^2}$

Answer 1

The correct option is B $\frac{{\mu }_{o}I}{3\pi a}\sqrt{4+{\pi }^2}$


Magnetic field at point $O$ due to different parts is as follows

Part $\left(1\right):$ $B_1=0$ [As lies on the same extended line]

Part $\left(2\right):$ ${\overrightarrow{B}}_2=\frac{{\mu }_{o}I}{2a}(-\hat{k})$

Part $\left(3\right):$ ${\overrightarrow{B}}_3=\frac{{\mu }_{o}I}{2\pi a}(-\hat{j})$

Part $\left(4\right):$ ${\overrightarrow{B}}_4=\frac{{\mu }_{o}I}{6a}\left(\hat{k}\right)$

Part $\left(5\right):$ ${\overrightarrow{B}}_5=\frac{{\mu }_{o}I}{6\pi a}(-\hat{j})$

$\therefore {\overrightarrow{B}}_2+{\overrightarrow{B}}_4=\frac{{\mu }_{o}I}{3a}(-\hat{k})$ and

${\overrightarrow{B}}_3+{\overrightarrow{B}}_5=\frac{2{\mu }_{o}I}{3\pi a}(-\hat{j})$

So, the net magnetic field,

$B_{net}=\sqrt{(|B_2+B_4|{)}^2+(|B_3+B_5|{)}^2}$

$=\frac{{\mu }_{o}I}{3\pi a}\sqrt{{\pi }^2+4}$

Hence, option (b) is the correct answer.