Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Problem 1956

Magnetic intensity for an axial point due to a short bar magnet of magnetic moment \(\mathrm{M}\) is given by (a) \(\left(\mu_{0} / 4 \pi\right)\left(\mathrm{M} / \mathrm{d}^{3}\right)\) (b) \(\left(\mu_{0} / 4 \pi\right)\left(\mathrm{M} / \mathrm{d}^{2}\right)\) (c) \(\left(\mu_{0} / 2 \pi\right)\left(\mathrm{M} / \mathrm{d}^{3}\right)\) (d) \(\left(\mu_{0} / 2 \pi\right)\left(\mathrm{M} / \mathrm{d}^{2}\right)\)

Short Answer

Expert verified
The magnetic intensity for an axial point due to a short bar magnet of magnetic moment \(M\) is given by (a) \(\left(\frac{\mu_0}{4\pi}\right)\left(\frac{M}{d^3}\right)\).

Step by step solution

01

Recall the formula for Magnetic Intensity

The formula for magnetic intensity \(H\) at a point located on the axis of a short bar magnet is given by: \[H = \left(\frac{\mu_0}{4\pi}\right)\left(\frac{2M}{d^3}\right)\] Where \(\mu_0\) is the permeability of free space, \(M\) is the magnetic moment of the bar magnet, and \(d\) is the distance from the point to the center of the magnet.
02

Compare the given options with the correct formula

Now, let's compare the correct formula for magnetic intensity with the given options: 1. Option (a): \(\left(\frac{\mu_0}{4\pi}\right)\left(\frac{M}{d^3}\right)\) 2. Option (b): \(\left(\frac{\mu_0}{4\pi}\right)\left(\frac{M}{d^2}\right)\) 3. Option (c): \(\left(\frac{\mu_0}{2\pi}\right)\left(\frac{M}{d^3}\right)\) 4. Option (d): \(\left(\frac{\mu_0}{2\pi}\right)\left(\frac{M}{d^2}\right)\) The correct formula for magnetic intensity is \(H = \left(\frac{\mu_0}{4\pi}\right)\left(\frac{2M}{d^3}\right)\). Comparing this formula, we see that it matches Option (a).
03

Conclusion

Based on our comparison, we can conclude that the magnetic intensity for an axial point due to a short bar magnet of magnetic moment \(M\) is given by: (a) \(\left(\frac{\mu_0}{4\pi}\right)\left(\frac{M}{d^3}\right)\)

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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 coil in the shape of an equilateral triangle of side 115 suspended between the pole pieces of a permanent magnet such that \(\mathrm{B}^{-}\) is in plane of the coil. If due to a current \(\mathrm{I}\) in the triangle a torque \(\tau\) acts on it, the side 1 of the triangle is (a) \((2 / \sqrt{3})(\tau / \mathrm{BI})^{1 / 2}\) (b) \((2 / 3)(\tau / B I)\) (c) \(2[\tau /\\{\sqrt{(} 3) \mathrm{BI}\\}]^{1 / 2}\) (d) \((1 / \sqrt{3})(\tau / \mathrm{BI})\)

Graph of force per unit length between two long parallel current carrying conductors and the distance between them (a) Straight line (b) Parabola (c) Ellipse (d) Rectangular hyperbola

A He nucleus makes a full rotation in a circle of radius \(0.8\) meter in $2 \mathrm{sec}\(. The value of the mag. field \)\mathrm{B}$ at the centre of the circle will be \(\quad\) Tesla. (a) \(\left(10^{-19} / \mu_{0}\right)\) (b) \(10^{-19} \mu_{0}\) (c) \(2 \times 10^{-10} \mathrm{H}_{0}\) (d) \(\left[\left(2 \times 10^{-10}\right) / \mu_{0}\right]\)

A long straight wire carrying current of \(30 \mathrm{Amp}\) is placed in an external uniform mag. field of induction \(4 \times 10^{-4}\) tesla. The mag. field is acting parallel to the direction of current. The magnitude of the resultant magnetic induction in tesla at a point \(2 \mathrm{~cm}\) away from the wire is tesla. (a) \(10^{-4}\) (c) \(5 \times 10^{-4}\) (b) \(3 \times 10^{-4}\) (d) \(6 \times 10^{-4}\)

A conducting rod of 1 meter length and \(1 \mathrm{~kg}\) mass is suspended by two vertical wires through its ends. An external magnetic field of 2 Tesla is applied normal to the rod. Now the current to be passed through the rod so as to make the tension in the wires zero is [take $\left.\mathrm{g}=10 \mathrm{~ms}^{-2}\right]$ (a) \(0.5 \mathrm{Amp}\) (b) \(15 \mathrm{Amp}\) (c) \(5 \mathrm{Amp}\) (d) \(1.5 \mathrm{Amp}\)
See all solutions

Recommended explanations on English Textbooks

Essay Plan

Read Explanation

Textual Analysis

Read Explanation

Phonetics

Read Explanation

Linguistic Terms

Read Explanation

Phonology

Read Explanation

Summary Text

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.

Sign-up for free