Question 6.25: Notice the following parallel: { ∇.D=0,∇×E=0, ε0E=D-P,(on fr...

Notice the following parallel:

\left\{\begin{array}{lll} \nabla \cdot D =0, & \nabla \times E = 0 , & \epsilon_{0} E = D - P , & \text { (no free charge); } \\\nabla \cdot B =0, & \nabla \times H = 0 , & \mu_{0} H = B -\mu_{0} M , & \text { (no free current). } \end{array}\right.

Thus, the transcription  D \rightarrow B , E \rightarrow H , P \rightarrow \mu_{0} M , \epsilon_{0} \rightarrow \mu_{0} turns an electrostatic problem into an analogous magnetostatic one. Use this, together with your knowledge of the electrostatic results, to rederive

(a) the magnetic field inside a uniformly magnetized sphere (Eq. 6.16);

B =\frac{2}{3} \mu_{0} M                           (6.16)

(b) the magnetic field inside a sphere of linear magnetic material in an otherwise uniform magnetic field (Prob. 6.18);

(c) the average magnetic field over a sphere, due to steady currents within the sphere (Eq. 5.93).

B _{ ave }=\frac{\mu_{0}}{4 \pi} \frac{2 m }{R^{3}}                                (5.93)

The Blue Check Mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

(a) The electric field inside a uniformly polarized sphere,  E =-\frac{1}{3 \epsilon_{0}} P (Eq. 4.14) translates to  H =-\frac{1}{3 \mu_{0}}\left(\mu_{0} M \right)=-\frac{1}{3} \text { M. } \operatorname{But} B =\mu_{0}( H + M ) .

E =-\nabla V=-\frac{P}{3 \epsilon_{0}} \hat{ z }=-\frac{1}{3 \epsilon_{0}} P , \quad \text { for } \quad r<R                         (4.14)

So the magnetic field inside a uniformly magnetized sphere is B =\mu_{0}\left(-\frac{1}{3} M + M \right)=\frac{2}{3} \mu_{0} M (same as Eq. 6.16).

B =\frac{2}{3} \mu_{0} M                             (6.16)

(b) The electric field inside a sphere of linear dielectric in an otherwise uniform electric field is E =\frac{1}{1+\chi_{e} / 3} E _{0} (Eq. 4.49). 

E =\frac{3}{\epsilon_{r}+2} E _{0}                             (4.49)

Now \chi_{e} \text { translates to } \chi_{m} , for then Eq. 4.30 \left( P =\epsilon_{0} \chi_{e} E \right) \text { goes to } \mu_{0} M =\mu_{0} \chi_{m} H , \text { or } M =\chi_{m} H (Eq. 6.29). So Eq. 4.49 \Rightarrow H =\frac{1}{1+\chi_{m} / 3} H _{0} \text {. But } B =\mu_{0}\left(1+\chi_{m}\right) H , \text { and } B _{0}=\mu_{0} H _{0} (Eqs. 6.31 and 6.32), so the magnetic field inside a sphere of linear magnetic material in an otherwise uniform magnetic field is

\frac{ B }{\mu_{0}\left(1+\chi_{m}\right)}=\frac{1}{\left(1+\chi_{m} / 3\right)} \frac{ B _{0}}{\mu_{0}} , or B =\left(\frac{1+\chi_{m}}{1+\chi_{m} / 3}\right) B _{0} (as in Prob. 6.18).

P =\epsilon_{0} \chi_{e} E                            (4.30)

M =\chi_{m} H                               (6.29)

B =\mu H                                   (6.31)

\mu \equiv \mu_{0}\left(1+\chi_{m}\right)                  (6.32)

(c) The average electric field over a sphere, due to charges within, is E _{\text {ave }}=-\frac{1}{4 \pi \epsilon_{0}} \frac{ p }{R^{3}} . Let’s pretend the charges are all due to the frozen-in polarization of some medium (whatever ρ might be, we can solve  \nabla \cdot P =-\rho to find the appropriate P). In this case there are no free charges, and p =\int P d \tau, \text { so } E _{\text {ave }}=-\frac{1}{4 \pi \epsilon_{0}} \frac{1}{R^{3}} \int P d \tau , which translates to

H _{ ave }=-\frac{1}{4 \pi \mu_{0}} \frac{1}{R^{3}} \int \mu_{0} M d \tau=-\frac{1}{4 \pi R^{3}} m .

But  B =\mu_{0}( H + M ), \text { so } B _{ ave }=-\frac{\mu_{0}}{4 \pi} \frac{ m }{R^{3}}+\mu_{0} M _{\text {ave }}, \text { and } M _{\text {ave }}=\frac{ m }{\frac{4}{3} \pi R^{3}}, \text { so } B _{\text {ave }}=\frac{\mu_{0}}{4 \pi} \frac{2 m }{R^{3}} , in agreement

with Eq. 5.93. (We must assume for this argument that all the currents are bound, but again it doesn’t really matter, since we can model any current configuration by an appropriate frozen-in magnetization. See G. H. Goedecke, Am. J. Phys. 66, 1010 (1998).)

B _{ ave }=\frac{\mu_{0}}{4 \pi} \frac{2 m }{R^{3}}                    (5.93)

Related Answered Questions