Q166 / 180 · COMEDK 2025 (10 May Shift 3) · Physics
A metal rod of susceptibility 799 is subjected to a magnetising field of $2000 \mathrm{Am}^{-1}$. The permeability of the material of the rod is: (Given $\mu_0=4 \pi \times 10^{-7} \mathrm{TmA}^{-1}$ )
- A$4.2 \pi \times 10^{-7} \mathrm{TmA}^{-1}$
- B$2.4 \pi \times 10^{-5} T m A^{-1}$
- Correct answerC$3.2 \pi \times 10^{-4} \mathrm{TmA}^{-1}$
- D$80 \pi \times 10^{-7} T m A^{-1}$
Solution
The magnetic susceptibility $\chi$ of the material is given as $799$.
The relative permeability $\mu_r$ is related to the magnetic susceptibility by the formula $\mu_r = 1 + \chi$.
Substituting the given value, $\mu_r = 1 + 799 = 800$.
The absolute permeability $\mu$ of the material is given by $\mu = \mu_0 \mu_r$.
Given $\mu_0 = 4 \pi \times 10^{-7} \text{ TmA}^{-1}$, we have $\mu = (4 \pi \times 10^{-7}) \times 800$.
Calculating the value, $\mu = 3200 \pi \times 10^{-7} \text{ TmA}^{-1} = 3.2 \pi \times 10^{-4} \text{ TmA}^{-1}$.
Answer: $3.2 \pi \times 10^{-4} \mathrm{TmA}^{-1}$
The relative permeability $\mu_r$ is related to the magnetic susceptibility by the formula $\mu_r = 1 + \chi$.
Substituting the given value, $\mu_r = 1 + 799 = 800$.
The absolute permeability $\mu$ of the material is given by $\mu = \mu_0 \mu_r$.
Given $\mu_0 = 4 \pi \times 10^{-7} \text{ TmA}^{-1}$, we have $\mu = (4 \pi \times 10^{-7}) \times 800$.
Calculating the value, $\mu = 3200 \pi \times 10^{-7} \text{ TmA}^{-1} = 3.2 \pi \times 10^{-4} \text{ TmA}^{-1}$.
Answer: $3.2 \pi \times 10^{-4} \mathrm{TmA}^{-1}$
Hence, the correct answer is option: 'C'.