Q73P A bar magnet whose magnetic dipo... [FREE SOLUTION]

Chapter 20: Q73P (page 863)

A bar magnet whose magnetic dipole moment is $15 {鈥碱.尘}^{2}$ is aligned with an applied magnetic field of $4 鈥涂$ . How much work must you do to rotate the bar magnet $180 掳$ to point in the direction opposite to the magnetic field?

Short Answer

Expert verified

The work done to rotate the bar magnet $180 掳$ to point in the direction opposite to the magnetic field is 120 J.

Step by step solution

Given Information

The dipole moment of the magnet is $渭 = 15 {鈥碱.尘}^{2}$ , and the magnetic field is $B = 4 鈥涂$ . T. The bar rotates at an angle $胃 = 180 掳$ .

Work done

The directions of the bar rotation and the magnetic torque are the same, therefore, the work done is non-zero and equals to the difference between the two potential energies at the angle $胃_{1} = 0$ and $胃_{2} = 180 掳$ given as:

$W = 螖 U_{2} 鈭� 螖 U_{1} 鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌� 鈰� (1)$

Where the potential energy for a magnetic dipole is given by

$\begin{matrix} U = 鈭� \overset{鈫�}{渭} 鈰� \overset{鈫�}{B} \\ = 鈭� 渭 B \cos 胃 \end{matrix}$

Therefore, the equation can be written as:

$\begin{matrix} W = 螖 U_{1} 鈭� 螖 U_{2} \\ = 鈭� 渭 B \cos 胃_{2} 鈭� (鈭� 渭 B \cos 胃_{1}) \\ = 渭 B (\cos 胃_{1} 鈭� \cos 胃_{2}) 鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆€� 鈰� (2) \end{matrix}$