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# A potential energy of a bar magnet moment $8A{m^2}$ placed in a uniform magnetic field of $2T$ at an angle of $120^\circ$ is equal toA) $- 16{\text{ }}J$B) $16{\text{ }}J$C) $- 8{\text{ }}J$D) $8{\text{ }}J$

Last updated date: 20th Jun 2024
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Hint: The potential energy of a bar magnet is proportional to the dot product of the magnetic moment and the uniform magnetic field. It is minimum when the magnet is perpendicular to the magnetic field and maximum when it is aligned with it.

Formula used:
In this solution, we will use the following formula:
Potential energy when a magnet is placed in a magnetic field: $U = - M.B$ where $M$ is the magnetic moment and $B$ is the external magnetic field.

We want to calculate the potential energy of a bar magnet when it is placed in an external magnetic field. We know that the potential energy is calculated as the dot product of the magnetic moment of the magnet and the strength of the external magnetic field. So, we can write the potential energy as
$U = - M.B$
In our case, the magnetic moment of the bar magnet is $8A{m^2}$ and the strength of the external magnetic field is $2T$ and the angle between these two vectors is given as $120^\circ$. SO, the potential energy will be
$U = - MB\cos 120^\circ$
So, substituting the value of $M = 8A{m^2}$ and $B = 2T$, we can calculate the potential energy as
$U = - 8 \times 2 \times \dfrac{{ - 1}}{2}$
Which gives us
$U = 8\,J$

Note: We must be careful to not forget the minus sign in the formula for potential energy. The potential energy of the magnet will be minimum when it is aligned with the external magnetic which means the system will always try to minimize the potential energy and, in this case, also, the magnet will move such that it aligns as described above.