JEE Main 2026 — Magnetic Effects of Current Question with Solution

For the first figure, the magnetic field at the centre P is the vector sum of the magnetic fields due to two semi-infinite straight wires and one semicircular arc.

The magnetic field due to the top semi-infinite wire at P is $B_{1a}=\frac{{\mu }_{0}I}{4\pi r}$ directed into the plane.

The magnetic field due to the bottom semi-infinite wire at P is $B_{1b}=\frac{{\mu }_{0}I}{4\pi r}$ directed into the plane.

The magnetic field due to the semicircular arc at P is $B_{1c}=\frac{{\mu }_{0}I}{4r}=\frac{\pi {\mu }_{0}I}{4\pi r}$ directed into the plane.

The total magnetic field at P is $B_1=B_{1a}+B_{1b}+B_{1c}=\frac{{\mu }_{0}I}{4\pi r}(1+1+\pi )=\frac{{\mu }_{0}I}{4\pi r}(2+\pi )$.

For the second figure, the magnetic field at the centre Q is the vector sum of the magnetic fields due to one horizontal semi-infinite wire, one semicircular arc, and one vertical semi-infinite wire.

The magnetic field due to the horizontal semi-infinite wire at Q is $B_{2a}=\frac{{\mu }_{0}I}{4\pi r}$ directed into the plane.

The magnetic field due to the semicircular arc at Q is $B_{2c}=\frac{{\mu }_{0}I}{4r}=\frac{\pi {\mu }_{0}I}{4\pi r}$ directed into the plane.

Since the arc is semicircular, it ends at a point directly below Q. The vertical semi-infinite wire starts from this point and extends downwards, meaning its line of action passes through Q. Therefore, the magnetic field due to this vertical wire at Q is $B_{2b}=0$.

The total magnetic field at Q is $B_2=B_{2a}+B_{2b}+B_{2c}=\frac{{\mu }_{0}I}{4\pi r}(1+0+\pi )=\frac{{\mu }_{0}I}{4\pi r}(1+\pi )$.

The ratio of the magnetic fields is $\frac{B_1}{B_2}=\frac{\frac{{\mu }_{0}I}{4\pi r}(2+\pi )}{\frac{{\mu }_{0}I}{4\pi r}(1+\pi )}=\frac{2+\pi }{1+\pi }$.

Answer: $\frac{2+\pi }{1+\pi }$