JEE Main 2026 — Magnetic Properties of Matter Question with Solution

Let the magnetic moment of each magnet be $M=3\sqrt{5}$ J/T.

The distance of point P from the center of each magnet is $r=5$ cm $=0.05$ m.

For the first magnet (horizontal), point P lies on its axial line. The magnetic field at P due to the first magnet is:

$B_1=\frac{{\mu }_0}{4\pi }\frac{2M}{r^3}$

The direction of ${\vec{B}}_1$ is along the axis of the first magnet.

For the second magnet (vertical), point P lies on its equatorial line. The magnetic field at P due to the second magnet is:

$B_2=\frac{{\mu }_0}{4\pi }\frac{M}{r^3}$

The direction of ${\vec{B}}_2$ is parallel to the axis of the second magnet, which is perpendicular to the axis of the first magnet.

Since ${\vec{B}}_1$ and ${\vec{B}}_2$ are perpendicular to each other, the net magnetic field at P is:

$B=\sqrt{B_1^2+B_2^2}$

Substituting the expressions for $B_1$ and $B_2$:

$B=\sqrt{{\left(\frac{{\mu }_0}{4\pi }\frac{2M}{r^3}\right)}^2+{\left(\frac{{\mu }_0}{4\pi }\frac{M}{r^3}\right)}^2}$

$B=\frac{{\mu }_0}{4\pi }\frac{M}{r^3}\sqrt{2^2+1^2}=\frac{{\mu }_0}{4\pi }\frac{\sqrt{5}M}{r^3}$

Substituting the given values $\frac{{\mu }_0}{4\pi }=10^{-7}$ T m/A, $M=3\sqrt{5}$ J/T, and $r=0.05$ m:

$B=10^{-7}\times \frac{\sqrt{5}\times 3\sqrt{5}}{(0.05{)}^3}$

$B=10^{-7}\times \frac{15}{125\times 10^{-6}}$

$B=\frac{15\times 10^{-1}}{125}$

$B=0.1\times 0.12$ T

$B=0.012$ T

$B=12\times 10^{-3}$ T

Comparing this with the given value $\alpha \times 10^{-3}$ T, we get $\alpha =12$.

Answer: $12$