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Physics Question 39 – JEE-MAIN 2026

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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 in SI units is a×105. The value of a is _______. (Free space permeability μ0=4π×10−7 SI units.)

Understand the relationship between magnetic field (B), magnetic intensity (H), and permeability (μ) in a material.

Step 1: Calculate the permeability of the material✦ Active

The permeability of the material (μ) is given by the product of its relative permeability (μr) and the permeability of free space (μ0). Given μr=400 and μ0=4π×10−7 T m/A.

μ=μrμ0=400×(4π×10−7)=1600π×10−7 T m/A
Step 2: Determine the magnetic intensity (H)○ Expand

The magnetic field (B) inside the solenoid is related to the magnetic intensity (H) by the material's permeability (μ). Given B=1.0 T.

B=μH⟹H=Bμ H=1.01600π×10−7=1071600π=10000016π=6250π A/m
Step 3: Find the value of a○ Expand

The magnetic intensity is given in the form a×105. Equate this to the calculated value of H to find a.

a×105=6250π a=6250π×105=6250100000π=62510000π=116π
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