Magnetic Properties of Matter

JEE Main level, single correct, medium

2 of 57

Question 2 of 57

A solenoid has a core made of material with relative permeability 400400. The magnetic field produced in the interior of solenoid is 1.01.0 T. The magnetic intensity in SI units is α×105\alpha \times 10^5. The value of α\alpha is ______.
(Free space permeability μ0=4π×10−7\mu_0=4\pi \times 10^{-7} SI units.)

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Answer: (B) 116π\dfrac{1}{16\pi}

The magnetic field BB inside the solenoid is related to the magnetic intensity HH by the formula:

B=μH=μ0μrHB = \mu H = \mu_0 \mu_r H

Rearranging for HH, we get:

H=Bμ0μrH = \dfrac{B}{\mu_0 \mu_r}

Substituting the given values (B=1.0B = 1.0 T, μr=400\mu_r = 400, and μ0=4π×10−7\mu_0 = 4\pi \times 10^{-7} T m/A):

H=1.04π×10−7×400H = \dfrac{1.0}{4\pi \times 10^{-7} \times 400}

H=1.01600π×10−7H = \dfrac{1.0}{1600\pi \times 10^{-7}}

H=116π×10−5H = \dfrac{1}{16\pi \times 10^{-5}}

H=116π×105H = \dfrac{1}{16\pi} \times 10^5 A/m

Comparing this with the given expression H=α×105H = \alpha \times 10^5, we find:

α=116π\alpha = \dfrac{1}{16\pi}

Answer: 116π\dfrac{1}{16\pi}

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