/*! 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 148 A bar magnet has a magnetic mome... [FREE SOLUTION] | 91影视

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Chapter 15: Problem 148

A bar magnet has a magnetic moment of \(2.5 \mathrm{JT}^{-1}\) and is placed in a magnetic field of \(0.2 \mathrm{~T}\). Work done in turning the magnet from parallel to antiparallel position relative to field direction is

Short Answer

Expert verified
The work done in turning the bar magnet from the parallel to the antiparallel position relative to the field direction is \(2.5 \mathrm{J}\).

Step by step solution

01

Write down the given values

We are given the magnetic moment of the bar magnet (\(M = 2.5 \mathrm{JT}^{-1}\)) and the magnetic field strength (\(B = 0.2 \mathrm{T}\)).
02

Determine the magnetic flux at parallel and antiparallel positions

In the parallel position, the angle between the magnetic moment and the magnetic field is 0, so the magnetic flux \(\Phi_P = MB \cos(0)\). In the antiparallel position, the angle is 180掳, so the magnetic flux, \(\Phi_A = MB \cos(180)\).
03

Calculate the change in magnetic flux

The change in magnetic flux \(\Delta B\) can be calculated as follows: \(\Delta B = \Phi_A - \Phi_P\)
04

Calculate the work done

Using the formula for work done in turning a magnetic moment in a magnetic field, we have: \(W = - M \Delta B\) Now let's perform the calculations to find the work done. #Solution#
05

Write down the given values

The magnetic moment of the bar magnet (\(M\)) is given as \(2.5 \mathrm{JT}^{-1}\), and the magnetic field strength (\(B\)) is given as \(0.2 \mathrm{T}\).
06

Determine the magnetic flux at parallel and antiparallel positions

For the parallel position, \(\Phi_P = 2.5 \times 0.2 \cos(0) = 0.5 \mathrm{J}\). For the antiparallel position, \(\Phi_A = 2.5 \times 0.2 \cos(180) = -0.5 \mathrm{J}\).
07

Calculate the change in magnetic flux

We calculate the change in magnetic flux as follows: \(\Delta B = \Phi_A - \Phi_P = -0.5 - 0.5 = -1 \mathrm{J}\)
08

Calculate the work done

Now, we can find the work done using the formula: \(W = - M \Delta B = - 2.5 \times -1 = 2.5 \mathrm{J}\) The work done in turning the bar magnet from the parallel to the antiparallel position relative to the field direction is \(2.5 \mathrm{J}\).

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Most popular questions from this chapter

A magnet makes 30 oscillations per minute at a plane where intensity is \(32 \mathrm{~T}\). At another place it takes \(1 \mathrm{~s}\) to complete one oscillation. The value of horizontal intensity at the second place is, (A) \(12.8 \mathrm{~T}\) (B) \(25.6 \mathrm{~T}\) (C) \(128 \mathrm{~T}\) (D) \(256 \mathrm{~T}\)

A charged particle moves in a uniform magnetic field of induction \(\vec{B}\) with a velocity \(\vec{v}\). The change in kinetic energy in the magnetic field is zero when the velocity \(\vec{v}\) is (A) parallel to \(\vec{B}\) (B) perpendicular to \(\vec{B}\) (C) at any angle to \(\vec{B}\) (D) None of these

A proton and an alpha-particle enter a uniform magnetic field with the same velocity. The period of rotation of the alpha particle will be (A) four times that of the proton. (B) two times that of the proton. (C) three times that of the proton. (D) same as that of the proton.

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A charged particle is released from rest in a region of steady and uniform electric and magnetic fields which are parallel to each other. The particle will move in a (A) Straight line (B) Circle (C) Helix (D) Cycloid
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