JEE Main 2026 Apr 4 Shift 2, Physics Q14: Magnetic Field due to Electric Current

JEE Main2026Apr 4, Shift 2Physics

A solenoid has a core made of material with relative permeability 400. The magnetic field produced in the interior of solenoid is 1.0 T. The magnetic intensity is SI units is $α \times 1 0^{5}$ . The value of $α$ is ________. (Free space permeability $μ_{0} = 4 π \times 1 0^{- 7}$ SI units).

A
$\frac{25}{π}$
B
$\frac{1}{16 π}$
C
$\frac{1}{π}$
D
$\frac{1}{4 π}$

Solution

Magnetic intensity $H = \frac{B}{μ} = \frac{B}{μ _{0} μ _{r}}$ .

$H = \frac{1}{400 \times 4 π \times 1 0 ^{- 7}} = \frac{1}{16 π \times 1 0 ^{- 5}} = \frac{1}{16 π} \times 1 0^{5} A/m$ .

$∴ α = \frac{1}{16 π}$ .

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JEE Main 2026 Apr 4 Shift 2, Physics Q14: Magnetic Field due to Electric Current

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