Question

In the figure shown the magnetic field at the point P is

• 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. none
• D. both

Answer 1

The correct option is C none

Solution :

First we will find out the magnetic field due to semicircular wires

Magnetic field due to smaller semicircle is $B_1=\frac{\mu I}{4\times \frac{a}{2}}$ $\Rightarrow$ $B_1=\frac{\mu I}{2a}$ (inside the plane since current is clockwise)

Magnetic field due to bigger semicircle is $B_2=\frac{\mu I}{4\times \frac{3a}{2}}$ $\Rightarrow$ $B_2=\frac{\mu I}{6a}$(outside the plane since current is anti clockwise)

Net Magnetic field due to semicircles is $B_1-B2$ = $\frac{\mu I}{2a}$ - $\frac{\mu I}{6a}$ $\Rightarrow$ $B_{1net}=\frac{\mu I}{3a}$(inside the plane)

Now we will find out the magnetic field due to straight wires

Magnetic due to outer wire is $B_3=\frac{\mu I}{2\pi \frac{3a}{2}}$ $\Rightarrow$ $B_3=\frac{\mu I}{3\pi }$ (Vertically downwards acc to flemings right hand rule)

Magnetic due to inner wire is $B_4=\frac{\mu I}{2\pi \frac{a}{2}}$ $\Rightarrow$ $B_4=\frac{\mu I}{\pi }$ (Vertically downwards acc to flemings right hand rule)

Net magnetic field due to straight wires $B_3+B_4$=$\frac{\mu I}{3\pi }$+$\frac{\mu I}{\pi }$ $\Rightarrow$ $B_{2net}=\frac{4\mu I}{3\pi }$ (Vertically downwards )

Total magnetic field $B_{Total}=\sqrt{B_{1net}^2+B_{2net}^2}$

On solving we get $B_{Total}=\frac{\mu i}{3\pi a}$ $\times \sqrt{16a^2+{\pi }^2}$

The Correct Opt =C