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A bar magnet of magnetic moment $M$ is aligned parallel to the direction of uniform magnetic field. What is the work done to align its magnetic moment (i) opposite to the field and ii) normal to the field.

seo-qna
Last updated date: 13th Jun 2024
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Answer
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Hint When the magnetic field is parallel to the magnetic dipole moment, the angle between them is zero, and when it is antiparallel, the angle is considered $180^\circ $. The potential energy of an object in a magnetic field is proportional to the magnetic dipole moment, the magnetic field and the angle between them. From here we can calculate the work done as the change in potential energy.
Formula used: In this solution we will be using the following formula;
$\Rightarrow U = - m \cdot B$ where $U$ is the potential energy of an object with a dipole moment of $m$in magnetic field, and $B$ is the magnetic flux density of the magnetic field.
$\Rightarrow W = \Delta U$ where $W$ is the work done and $\Delta U$ is change in potential energy of the system.

Complete step by step answer
An isolated magnet has a magnetic dipole moment due to orbiting of the electrons about the nucleus. Such magnet now placed in a magnetic field, will possess a potential energy given by
$\Rightarrow U = - m \cdot B = - mB\cos \theta $ where $m$ is the magnetic moment of the magnet, $B$ is the magnetic flux density of the magnetic field, and $\theta $ is the angle between the magnetic moment and the magnetic field.
Hence for the magnet in our question, we have
$\Rightarrow U = - MB\cos \theta $ since the magnetic moment and the magnetic field are parallel
Then
$\Rightarrow U = - MB\cos 0^\circ = - MB$
Work done in a field is defined as the change in potential energy
$\Rightarrow W = \Delta U = {U_f} - {U_i}$
Now for work done in aligning its magnetic moment opposite to the field, we have
$\Rightarrow W = - MB\cos 180^\circ - \left( { - MB} \right)$(since opposite the field means anti-parallel thus $\theta = 180^\circ $). Hence,
$\Rightarrow W = MB - \left( { - MB} \right) = MB + MB$
$\therefore W = 2MB$
To align the magnetic moment to become normal to the magnetic field we have that
$\Rightarrow W = - MB\cos 90^\circ - ( - MB)$
Hence, by calculation
$\Rightarrow W = 0 - \left( { - MB} \right)$
$\Rightarrow W = MB$.

Note
This is the ideal work done which only considers the field. In reality, work will be done to overcome reactive forces such as friction or air resistance, which will thus increase the total amount of work done to move the magnet.