Q.61P Question: Using Eq. 5.88, calcul... [FREE SOLUTION]

91影视

Introduction to Electrodynamics

David J. Griffiths

Physics

4th edition Edition 路 613 pages

ISBN: 9780321856562

Chapter 5: Q.61P (page 264)

Question: Using Eq. 5.88, calculate the average magnetic field of a dipole over
a sphere of radius Rcentered 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 . (If you get stuck, refer to Prob. 3.48.) Evidently the truefield of a magnetic dipole is
${\overset{鈫�}{B}}_{dip} (\overset{鈫�}{r}) = \frac{渭_{0}}{4 {蟺谤}^{3}} [3 (\overset{鈫�}{m} 路 \hat{r}) \hat{r} - \overset{鈫�}{m}] + \frac{2 渭_{0}}{3} \overset{鈫�}{m} 未^{3} (\overset{鈫�}{r})$ ${\overset{鈫�}{B}}_{d i p} (\overset{鈫�}{r}) = \frac{渭_{0}}{4 蟺 r^{3}} [3 (\overset{鈫�}{m} 路 \hat{r}) \hat{r} - \overset{鈫�}{m}] + \frac{2 渭_{0}}{3} \overset{鈫�}{m} 未^{3} (\overset{鈫�}{r})$
Compare the electrostatic analog, Eq. 3.106.

Short Answer

Expert verified

Answer:

The average magnetic field of a dipole over a sphere of radius Rcentered at the origin is $\frac{渭_{0}}{4 蟺 r^{3}} [3 (\overset{鈫�}{m} 路 \hat{r}) \hat{r} - \overset{鈫�}{m}] + \frac{2 渭_{0}}{3} \overset{鈫�}{m} 未^{3} (\overset{鈫�}{r})$ $\frac{渭_{0}}{4 蟺 r^{3}} [3 (\overset{鈫�}{m} 路 \hat{r}) \hat{r} - \overset{鈫�}{m}] + \frac{2 渭_{0}}{3} \overset{鈫�}{m} 未^{3} (\overset{鈫�}{r})$ .

Step by step solution

Given data

A dipole having dipole moment $\overset{鈫�}{m}$ .

Magnetic field far from the dipole

The magnetic field outside an infinitesimal sphere centered at a dipole is

${\overset{鈫�}{B}}_{d i p} (\overset{鈫�}{r}) = \frac{渭_{0}}{4 蟺 r^{3}} [3 (\overset{鈫�}{m} 路 \hat{r}) \hat{r} - \overset{鈫�}{m}]$ ${\overset{鈫�}{B}}_{d i p} (\overset{鈫�}{r}) = \frac{渭_{0}}{4 蟺 r^{3}} [3 (\overset{鈫�}{m} 路 \hat{r}) \hat{r} - \overset{鈫�}{m}]$ 鈥�.. (1)

Here, $渭_{0}$ is the permeability of free space.

Magnetic field of a dipole

Inside the sphere, the magnetic field is a delta function

$\overset{鈫�}{B} = \overset{鈫�}{A} 未^{3} (\overset{鈫�}{r})$

Thus, the average field inside the sphere is

${\overset{鈫�}{B}}_{ave} = \frac{1}{\frac{4}{3} 蟺 R^{3}} 鈭� \overset{鈫�}{A} 未^{3} (\overset{鈫�}{r}) d 蟿 = \frac{3}{4 蟺 R^{3}} \overset{鈫�}{A}$

But the average field is also

${\overset{鈫�}{B}}_{ave} = \frac{渭_{0}}{4 蟺} \frac{2 \overset{鈫�}{m}}{R^{3}}$

Compare the two and get

$\overset{鈫�}{A} = \frac{2 渭_{0} \overset{鈫�}{m}}{3}$ $\overset{鈫�}{A} = \frac{2 渭_{0} \overset{鈫�}{m}}{3}$

Thus, the field inside the sphere is

role="math" localid="1658231587685" $\overset{鈫�}{B} = \frac{2 渭_{0} \overset{鈫�}{m}}{3} 未^{3} (\overset{鈫�}{r})$

From equation (1), the total field is

role="math" localid="1658231606913" ${\overset{鈫�}{B}}_{d i p} (\overset{鈫�}{r}) = \frac{渭_{0}}{4 蟺 r^{3}} [3 (\overset{鈫�}{m} 路 \hat{r}) \hat{r} - \overset{鈫�}{m}] + \frac{2 渭_{0} \overset{鈫�}{m}}{3} 未^{3} (\overset{鈫�}{r})$

Thus, the field of the dipole is $\frac{渭_{0}}{4 蟺 r^{3}} [3 (\overset{鈫�}{m} 路 \hat{r}) \hat{r} - \overset{鈫�}{m}] + \frac{2 渭_{0} \overset{鈫�}{m}}{3} 未^{3} (\overset{鈫�}{r})$ .