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Chapter 15: Problem 73

The magnetic susceptibility of a material of a rod is 499. Permeability of vacuum is \(4 \pi \times 10^{-7} \mathrm{H} / \mathrm{m}\). Absolute permeability of the material of the rod in henry per metre is (A) \(\pi \times 10^{-4}\) (B) \(2 \pi \times 10^{-4}\) (C) \(3 \pi \times 10^{-4}\) (D) \(4 \pi \times 10^{-4}\)

Short Answer

Expert verified
The absolute permeability of the material of the rod is \(2\pi \times 10^{-4} \, \mathrm{H} / \mathrm{m}\). So, the correct answer is (B).

Step by step solution

01

List the known values

We are given: - Magnetic susceptibility, \( \chi = 499 \) - Permeability of vacuum, \( \mu_0 = 4 \pi \times 10^{-7} \mathrm{H} / \mathrm{m} \)
02

Use the formula to find the absolute permeability

We need to use the formula \( \mu = \mu_0 (1 + \chi) \) to find the absolute permeability. Plug in the known values: \( \mu = (4 \pi \times 10^{-7}) (1 + 499) \)
03

Calculate \(\mu\)

First, calculate the value inside the brackets: \( 1 + 499 = 500 \) Now multiply by \(4 \pi \times 10^{-7}\): \( \mu = (500)(4\pi \times 10^{-7}) \) When we multiply, we get: \( \mu = 2\pi \times 10^{-4} \, \mathrm{H} / \mathrm{m} \)
04

Choose the correct option

Now, we have found the value of absolute permeability, and it matches with option (B): \(2\pi \times 10^{-4} \, \mathrm{H} / \mathrm{m}\) Hence, the correct answer is (B) \( 2\pi \times 10^{-4} \).

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Most popular questions from this chapter

A uniform horizontal magnetic field \(\vec{B}\) exists in the region. A rectangular loop of mass \(m\), horizontal side \(l\) (perpendicular to the magnetic field), and resistance \(R\) is placed in the region. The velocity with which it should be pushed down so that it continues to fall without acceleration will be (A) \(\frac{m g R}{B^{2} l^{2}}\) (B) \(\frac{B^{2} l^{2}}{m g R}\) (C) \(\frac{m g}{B l R}\) (D) \(\frac{R B l}{m g}\)

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