Question

A bar magnet of magnetic moment \[{10^4}J/T\;\] is free to rotate in a horizontal plane. The work done in rotating the magnet slowly from a direction parallel to a horizontal magnetic field of \[4 \times {10^{ - 5}}T\;\] to a direction \[\;{60^{o\;}}\]from the field will be
A) \[0.2J\]
B) \[2.0J\]
C) \[4.18J\]
D) \[2 \times {10^2}J\]

Answer 1

Hint:
We can use the formula to calculate the work performed in rotating a bar magnet through a certain angle in a magnetic field. A bar magnet's magnetic moment is parallel to the magnetic field when it is in a stable equilibrium position and in the final position bar magnet’s is ${60^0}$ with the magnetic field.



Complete step by step solution:
The bar magnet has a magnetic moment. The magnetic moment is a vector quantity that is the product of the magnitude of the magnetic field and the length of the magnetic moment arm. The magnetic moment is also the vector cross product of the magnetic field and the displacement vector from the North Pole of the magnet to the point in question.
We can use the formula directly. The external work done on a bar magnet rotating in a magnetic field depends on the magnitude and orientation of that magnet's magnetic moment in the initial and final positions.

\[W = MB\left( {cos{\theta _i} - cos{\theta _f}} \right){\text{ }} - - - - - - - (1)\;\]
where $W$= Work done, $M = $Magnetic moment, $B = $Magnetic field, ${\theta _i}{\text{ and }}{\theta _f}$are the initial and final positions of the bar magnet with the magnetic field.
The magnet is at its stable equilibrium position when it is in its initial orientation. The magnetic moment and magnetic field are parallel at the stable equilibrium position.
Hence, the magnetic moment's direction and the magnetic field's direction form an angle of ${\theta _i} = {0^0}$.
In the final orientation, the magnet is made with a ${60^0}$ with the magnetic field. Hence, the magnetic moment's direction and the magnetic field's direction form an angle of ${\theta _f} = {60^0}$. Let W represent the amount of effort required to rotate the bar magnet from its initial orientation to its final orientation in the magnetic field. Then we can write
 \[W = MB\left( {cos\left( {{0^0}} \right) - cos\left( {{{60}^0}} \right)} \right)J\]
\[\therefore W = MB\left( {1 - \left( {1/2} \right)} \right) = MB\left( {1/2} \right) = \dfrac{{MB}}{2}\]
Or, \[\therefore W = \dfrac{{MB}}{2} = \dfrac{{{{10}^4} \times 4 \times {{10}^{ - 5}}}}{2} = 0.2J\]
Therefore, the required work done is \[0.2J\] .
Hence, the correct option is A.




Therefore, the correct option is A.




Note:
It is important for students to remember that because electricity and magnetism are closely connected, a comparable equivalent formula exists for the effort required to spin an electric dipole through an angle in an electric field. The magnetic moment in formula (1) can be replaced with the magnitude of the electric dipole moment, and the electric field in formula (1) can be replaced with the size of the electric field. Therefore, students may also answer issues involving electric dipoles by just learning one of the formulas.