A magnetizing field of 10^3\ A/m produces a magnetic flux density 4π T in an iron rod. The relative permeability of the rod will be (permeability of free space μ_0 = 4π × 10^ -7 SI units)

Correct answer: A. 10^4 Given H = 10^3\ A/m and B = 4π\ T . The magnetic flux density is given by B = μ H = μ_0 μ_r H . Substituting the given values: 4π = (4π × 10^ -7 ) × μ_r × 10^3 μ_r = 4π 4π × 10^ -4 μ_r = 10^4 Answer: 10^4

MHT CET 202626 AprilMorning Shift

Exam: MHT CET
Year: 2026
Date: 26 April
Shift: Morning Shift
Subject: Physics
Chapter: Magnetic Properties of Matter

MHT CET 2026 · 26 April · Morning Shift · Physics · Magnetic Properties of Matter

A magnetizing field of 10^3\ A/m produces a magnetic flux density...

A magnetizing field of

10^3\ \text{A/m}

produces a magnetic flux density

4\pi

T in an iron rod. The relative permeability of the rod will be (permeability of free space

\mu_0 = 4\pi \times 10^{-7}

SI units)

Options

A $10^4$
B $10^5$
C $10^3$
D $10^6$

Correct answer

A. $10^4$

Step-by-step solution

Given

H = 10^3\ \text{A/m}

and

B = 4\pi\ \text{T}

.

The magnetic flux density is given by

B = \mu H = \mu_0 \mu_r H

.

Substituting the given values:

4\pi = (4\pi \times 10^{-7}) \times \mu_r \times 10^3

\mu_r = \dfrac{4\pi}{4\pi \times 10^{-4}}

\mu_r = 10^4

Answer:

10^4

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