Question 5.12: Use the result of Ex. 5.6 to calculate the magnetic field at...

Field (at center of sphere) due to the ring at \theta (see figure) is (Eq. 5.41):

B(z)=\frac{\mu_{0} I}{4 \pi}\left(\frac{\cos \theta}{ᴫ^{2}}\right) 2 \pi R=\frac{\mu_{0} I}{2} \frac{R^{2}}{\left(R^{2}+z^{2}\right)^{3 / 2}}                                    (5.41)

d B=\frac{\mu_{0} d I}{2} \frac{(R \sin \theta)^{2}}{\left[(R \sin \theta)^{2}+(R \cos \theta)^{2}\right]^{3 / 2}}=\frac{\mu_{0}}{2 R} \sin ^{2} \theta d I .

d I=K R d \theta, \quad K=\sigma v, \quad \sigma=\frac{Q}{4 \pi R^{2}}, \quad v=\omega R \sin \theta ,

so

d I=\frac{Q}{4 \pi R^{2}} \omega R \sin \theta R d \theta=\frac{Q \omega}{4 \pi} \sin \theta d \theta .

B=\frac{\mu_{0}}{2 R} \frac{Q \omega}{4 \pi} \int_{0}^{\pi} \sin ^{3} \theta d \theta=\frac{\mu_{0} Q \omega}{8 \pi R}\left(\frac{4}{3}\right) . \quad B =\frac{\mu_{0} Q \omega}{6 \pi R} \hat{ z } .