JEE Main 2026 — Physics Electromagnetism

Let the magnetic moment of each magnet be $M = 3\sqrt{5}$M=35​ J/T.

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

$$B_1 = \dfrac{\mu_0}{4\pi} \dfrac{2M}{r^3}$$B1​=4πμ0​​r32M​

The direction of $\vec{B}_1$B1​ 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:

$$B_2 = \dfrac{\mu_0}{4\pi} \dfrac{M}{r^3}$$B2​=4πμ0​​r3M​

The direction of $\vec{B}_2$B2​ is parallel to the axis of the second magnet, which is perpendicular to the axis of the first magnet.

Since $\vec{B}_1$B1​ and $\vec{B}_2$B2​ are perpendicular to each other, the net magnetic field at P is:

$$B = \sqrt{B_1^2 + B_2^2}$$B=B12​+B22​​

Substituting the expressions for $B_1$B1​ and $B_2$B2​:

$$B = \sqrt{\left(\dfrac{\mu_0}{4\pi} \dfrac{2M}{r^3}\right)^2 + \left(\dfrac{\mu_0}{4\pi} \dfrac{M}{r^3}\right)^2}$$B=(4πμ0​​r32M​)2+(4πμ0​​r3M​)2​

$$B = \dfrac{\mu_0}{4\pi} \dfrac{M}{r^3} \sqrt{2^2 + 1^2} = \dfrac{\mu_0}{4\pi} \dfrac{\sqrt{5}M}{r^3}$$B=4πμ0​​r3M​22+12​=4πμ0​​r35​M​

Substituting the given values $\dfrac{\mu_0}{4\pi} = 10^{-7}$4πμ0​​=10−7 T m/A, $M = 3\sqrt{5}$M=35​ J/T, and $r = 0.05$r=0.05 m:

$$B = 10^{-7} \times \dfrac{\sqrt{5} \times 3\sqrt{5}}{(0.05)^3}$$B=10−7×(0.05)35​×35​​

$$B = 10^{-7} \times \dfrac{15}{125 \times 10^{-6}}$$B=10−7×125×10−615​

$$B = \dfrac{15 \times 10^{-1}}{125}$$B=12515×10−1​

$B = 0.1 \times 0.12$B=0.1×0.12 T

$B = 0.012$B=0.012 T

$B = 12 \times 10^{-3}$B=12×10−3 T

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

Answer: $12$12