Question

An iron rod of susceptibility 599 is subjected to a magnetising field of 1200 A $m^{-1}$. The permeability of the material of the rod is : $(\mu_o = 4\pi \times 10^{-7} T\,m\,A^{-1})$

• A. $2.4 \pi \times 10^{-4} T\,m\,A^{-1}$
• B. $8.0 \pi \times 10^{-5} T\,m\,A^{-1}$
• C. $2.4 \pi \times 10^{-5} T\,m\,A^{-1}$
• D. $2.4 \pi \times 10^{-7} T\,m\,A^{-1}$

Answer 1

The Correct Option is A

To determine the permeability of the material of the rod, we can use the relationship between susceptibility (\(\chi_m\)) and permeability (\(\mu\)) in materials. The formula that relates magnetic susceptibility, permeability, and the permeability of free space (\(\mu_0\)) is given by:

\(\mu = \mu_0 (1 + \chi_m)\)

Where:

• \(\mu_0 = 4\pi \times 10^{-7} \, \text{T m/A}\) is the permeability of free space.
• \(\chi_m = 599\) is the magnetic susceptibility of the material.

Substituting the given values into the formula:

\(\mu = 4\pi \times 10^{-7} \times (1 + 599)\)

Simplifying the expression inside the parenthesis:

\(\mu = 4\pi \times 10^{-7} \times 600\)

Now, calculate the value:

\(\mu = 4\pi \times 600 \times 10^{-7}\)

\(\mu = 2400\pi \times 10^{-7}\)

Re-arranging the terms gives:

\(\mu = 2.4\pi \times 10^{-4} \, \text{T m/A}\)

Therefore, the permeability of the material of the rod is \(2.4 \pi \times 10^{-4} \, \text{T m/A}\).

The correct answer is \(2.4 \pi \times 10^{-4} \, \text{T m/A}\).