JEE Main 2026 Apr 4 Shift 2, Physics Q23: Magnetism and Matter

JEE Main2026Apr 4, Shift 2Physics

Two identical small bar magnets each of dipole moment $35 J/T$ 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 $α \times 1 0^{- 3} T$ . The value of $α$ is ________. ( $μ_{0} = 4 π \times 1 0^{- 7} Tm/A$ )

Your answer

Solution

P is at distance $r = 5 cm = 0.05 m$ from each magnet's centre.

From the axial magnet (P on its axis): $B_{1} = \frac{μ _{0}}{4 π} \cdot \frac{2 M}{r ^{3}}$ .

From the equatorial magnet (P on its equator): $B_{2} = \frac{μ _{0}}{4 π} \cdot \frac{M}{r ^{3}}$ .

The two field vectors are mutually perpendicular (one along the axial magnet's axis, the other antiparallel to the equatorial magnet's dipole). So:

$B_{n e t} = B_{1}^{2} + B_{2}^{2} = \frac{μ _{0}}{4 π} \cdot \frac{M}{r ^{3}} 4 + 1 = \frac{μ _{0}}{4 π} \cdot \frac{M 5}{r ^{3}}$ .

Substituting $M = 35$ , $r^{3} = (0.05)^{3} = 1.25 \times 1 0^{- 4} m^{3}$ :

$B_{n e t} = 1 0^{- 7} \cdot \frac{3 5 \cdot 5}{1.25 \times 1 0 ^{- 4}} = 1 0^{- 7} \cdot \frac{15}{1.25 \times 1 0 ^{- 4}} = 1 0^{- 7} \cdot 1.2 \times 1 0^{5} = 1.2 \times 1 0^{- 2} T$ .

$B_{n e t} = 12 \times 1 0^{- 3} T \Rightarrow α = 12$ .

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JEE Main 2026 Apr 4 Shift 2, Physics Q23: Magnetism and Matter

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