/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 46 The magnetic susceptibility of a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Problem 46

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 meter 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 short answer is: The absolute permeability of the rod material is (B) \(2 \pi \times 10^{-4} \mathrm{H} / \mathrm{m}\).

Step by step solution

01

Identify given values and formula

We are given: Magnetic susceptibility (\(\chi_m\)) = 499 Permeability of vacuum (\(\mu_0\)) = \(4 \pi \times 10^{-7} \mathrm{H} / \mathrm{m}\) We need to use the formula: \(\mu_{\text{material}} = \mu_0(1 + \chi_m)\)
02

Plug in the given values into the formula

Substitute the given values into the formula: \(\mu_{\text{material}} = (4 \pi \times 10^{-7})(1 + 499)\)
03

Calculate the absolute permeability

Simplify the equation: \(\mu_{\text{material}} = (4 \pi \times 10^{-7})(500)\) \(\mu_{\text{material}} = 2 \pi \times 10^{-4}\)
04

Compare the result to the given options

We can now see that our result matches option (B): \(\mu_{\text{material}} = 2 \pi \times 10^{-4}\) So, the correct answer is (B) \(2 \pi \times 10^{-4}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A particle of mass \(M\) and charge \(Q\) moving with a velocity \(\vec{v}\) describes a circular path of radius \(R\) when subjected to a uniform transverse magnetic field of induction \(B\). The work done by the field when the particle completes a full circle is (A) Zero (B) \(B Q 2 \pi R\) (C) \(B Q v(2 \pi R)\) (D) \(\left(\frac{M v^{2}}{R}\right)(2 \pi R)\)

In a transformer, number of turns in primary coil are 140 and that of the secondary coil are 280 . If current in primary coil is \(4 \mathrm{~A}\), then that of the secondary coil is (A) \(4 \mathrm{~A}\) (B) \(2 \mathrm{~A}\) (C) \(6 \mathrm{~A}\) (D) \(10 \mathrm{~A}\)

A coil of inductance \(1 H\) and resistance \(10 \Omega\) is connected to a resistance-less battery of EMF \(50 \mathrm{~V}\) at time \(t=0 .\) The ratio of rate at which magnetic energy is stored in the coil to the rate at which energy is supplied by the battery at \(t=0.1 \mathrm{~s} .\) is \(x \times 10^{-2}\). Find the value of \(x\). (Given \(\left.\frac{1}{e}=0.37\right)\)

A uniform conducting ring of mass \(\pi \mathrm{kg}\) and radius \(1 \mathrm{~m}\) is kept on smooth horizontal table. A uniform but time varying magnetic field \(B=\left(\hat{i}+t^{2} \hat{j}\right)\) Tesla is present in the region. (Where \(t\) is time in seconds). Resistance of ring is \(2 \Omega .\) Then, \(\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) Time (in second) at which ring start toppling is (A) \(\frac{10}{\pi}\) (B) \(\frac{20}{\pi}\) (C) \(\frac{5}{\pi}\) (D) \(\frac{25}{\pi}\)

There is a current of \(1.344 \mathrm{~A}\) in a copper wire whose area of cross-section normal to the length of the wire is \(1 \mathrm{~mm}^{2}\). If the number of free electrons per \(\mathrm{cm}^{3}\) is \(8.4 \times 10^{22}\), then the drift velocity of electrons will be (A) \(1.0 \mathrm{~mm} \mathrm{~s}\) (B) \(1.0\) meter s (C) \(0.1 \mathrm{~mm} \mathrm{~s}\) (D) \(0.01 \mathrm{~mm} \mathrm{~s}\)
See all solutions

Recommended explanations on Physics Textbooks

Scientific Method Physics

Read Explanation

Wave Optics

Read Explanation

Solid State Physics

Read Explanation

Torque and Rotational Motion

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Conservation of Energy and Momentum

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.