Two identical small bar magnets each of dipole moment 3√(5) J/T are… | JEE Main 2026 (4 Apr, Shift 2)

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Field of a Magnet & Magnet in a Uniform Field

Two identical small bar magnets each of dipole moment $3\sqrt{5}\text{ J/T}$ are placed at a center to center separation of 10 cm, with their axes perpendicular to each other as shown in figure. The value of magnetic field at the point P midway between the magnets is $\alpha \times 10^{-3}\text{ T}$ . The value of $\alpha$ is $\underline{\qquad}\ \left(\mu_0 = 4\pi \times 10^{-7}\text{ Tm/A}\right)$

Numerical answer type. Work it out, then open the answer.

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Answer: 12

P is 5 cm from each centre: on the axis of the left magnet and on the equatorial line of the right magnet.
$B_1 = \dfrac{\mu_0}{4\pi}\dfrac{2M}{r^3}$ (horizontal), $B_2 = \dfrac{\mu_0}{4\pi}\dfrac{M}{r^3}$ (vertical); they are perpendicular.
$B = \dfrac{\mu_0}{4\pi}\dfrac{\sqrt{5}M}{r^3} = 10^{-7} \times \dfrac{\sqrt{5} \times 3\sqrt{5}}{(0.05)^3} = \dfrac{15 \times 10^{-7}}{1.25 \times 10^{-4}}$
$B = 1.2 \times 10^{-2}\ \text{T} = 12 \times 10^{-3}\ \text{T}$
So $\alpha = 12$ .

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