Using Eq. 5.88, calculate the average magnetic field of a dipole over a sphere of radius R centered at the origin. Do the angular integrals first. Compare your answer with the general theorem in Prob. 5.59. Explain the discrepancy, and indicate how Eq. 5.89 can be corrected to resolve the

Question

Using Eq. 5.88, calculate the average magnetic field of a dipole over a sphere of radius R centered at the origin. Do the angular integrals first. Compare your answer with the general theorem in Prob. 5.59. Explain the discrepancy, and indicate how Eq. 5.89 can be corrected to resolve the ambiguity at r = 0. (If you get stuck, refer to Prob. 3.48.)

[latex]B _{ ext {dip }}( r )=\frac{\mu_{0}}{4 \pi} \frac{1}{r^{3}}[3( m \cdot \hat{ r }) \hat{ r }- m ][/latex] (5.89)

Evidently the true field of a magnetic dipole i[latex]s ^{29}[/latex]

[latex]B _{ dip }( r )=\frac{\mu_{0}}{4 \pi} \frac{1}{r^{3}}[3( m \cdot \hat{ r }) \hat{ r }- m ]+\frac{2 \mu_{0}}{3} m \delta^{3}( r )[/latex] (5.94)

Compare the electrostatic analog, Eq. 3.106.

[latex]E _{ ext {dip }}( r )=\frac{1}{4 \pi \epsilon_{0}} \frac{1}{r^{3}}[3( p \cdot \hat{ r }) \hat{ r }- p ]-\frac{1}{3 \epsilon_{0}} p \delta^{3}( r )[/latex] (3.106)

Accepted Answer

The issue (and the integral) is identical to the one in Prob. 3.48. The resolution (as before) is to regard Eq. 5.89 as correct outside an infinitesimal sphere centered at the dipole. Inside this sphere the field is a delta-function, [latex]A \delta^{3}( r )[/latex] , with A selected so as to make the average field consistent with Prob. 5.59:

[latex]B _{ ave }=\frac{1}{(4 / 3) \pi R^{3}} \int A \delta^{3}( r ) d au=\frac{3}{4 \pi R^{3}} A =\frac{\mu_{0}}{4 \pi} \frac{2 m }{R^{3}} \Rightarrow A =\frac{2 \mu_{0} m }{3}[/latex] . The added term is [latex] \frac{2 \mu_{0}}{3} m \delta^{3}( r )[/latex] .

Question 5.61: Using Eq. 5.88, calculate the average magnetic field of a di...

Related Answered Questions