A bar magnet whose magnetic dipole moment is 15 A ∗ m 2 is aligned with an applied magnetic field of 4 T. How much work must you do to rotate the bar magnet 180 ∘ to point in the direction opposite to the magnetic field?

Question

A bar magnet whose magnetic dipole moment is   15  A  ∗      m          2       is aligned with an applied magnetic field of 4 T. How much work must you do to rotate the bar magnet       180          ∘       to point in the direction opposite to the magnetic field?

Madilyn Dunn · Accepted Answer

An aligned magnetic dipole is associated with lower potential energy in the magnetic field, and the system tries to go to the lower potential-energy configuration. The directions of the magnetic torque and the bar rotation are the same, so the work done is nonzero and equals the difference between the two potential energies at the angle       θ          1        =  0 and       θ          2        =      180          ∘        W  =  △      U          2        −  △      U          1      Where the potential energy for a magnetic dipole is given by  U  =  −      μ    →    ∗      B    →    =  −  μ  B  cos  ⁡  θ ...(1)Therefore, equation (1) will be in the form  W  =  △      U          1        −  △      U          2        =  −  μ  B  cos  ⁡      θ          2        −  (  −  μ  B  cos  ⁡      θ          1        )  =  μ  B  (  cos  ⁡      θ          1        −  cos  ⁡      θ          2        ) ...(2)Now we can plug our values for   μ  ,  B  ,      θ          1       and       θ          2       into equation (2) to get the work done  W  =  μ  B  (  cos  ⁡      θ          1        −  cos  ⁡      θ          2        )  =  (  15  A  ∗      M          2        )  (  4  T  )  (  cos  ⁡      0          ∘        −  cos  ⁡      180          ∘        )=(60J)(1-(-1))=120JResult:W=120J

A bar magnet whose magnetic dipole moment is 15A*m^2 is aligned with an applied magnetic field of 4 T. How much work must you do to rotate the bar magnet 180^o to point in the direction opposite to the magnetic field?

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