Question

A uniformly magnetized (M -M) extruded ring, as shown, has inner diameter d and outer diameter D with thickness I. Calculate the magnetic field at the position of the center of the ring. Use Ampere's law to justify your result in the limit ld, D D d

Answer 1

If Instead of extruded ring with constant magnetization, then there is an object with a magnetic pole density is given as -

_$\rho$m = - _{$\bigtriangledown$}. $\vec{M}$ = 0

And a surface density of magnetic pole strength is given as -

$\sigma$_m = $\vec{M}$ . $\hat{n}$

To calculate the magnetic scalar potential (_$\phi$*) and ($\vec{B}$) from given equations as :

_$\phi$* = (1 / 4$\pi$) _$\int$$\sigma$_m dA' / | $\Gamma$ - $\Gamma$' |

$\vec{B}$ = - _{$\mu$0$\bigtriangledown$$\phi$}*

By symmetry, $\vec{B}$ on axis must be given as -

$\vec{B}$ = B_z (z) $\hat{z}$

$\vec{B}$ = - _$\mu$0 (d_$\phi$* / dz) $\hat{z}$                                                                   { eq.1 }

Using cylindrical coordinates (_$\rho$, $\Theta$, z), then we have

_$\phi$* _{(z, $\rho$=0)} = _$\phi$*₁ + _$\phi$*₂

_$\phi$* _{(z, $\rho$=0)} = (1 / 4$\pi$) _$\int$(M _$\rho$' d_$\rho$' d$\Theta$') / [_$\rho$^,2 + (z - l/2)²]^1/2 + (1 / 4$\pi$) _$\int$-(M _$\rho$' d_$\rho$' d$\Theta$') / [_$\rho$^,2 + (z + l/2)²]^1/2

_$\phi$* _{(z, $\rho$=0)} = (M/2) _{$\int_{d/2}^{D/2}$ $\rho$}' d_$\rho$' / [_$\rho$^,2 + (z - l/2)²]^1/2 - (M/2) _{$\int_{d/2}^{D/2}$ $\rho$}' d_$\rho$' / [_$\rho$^,2 + (z + l/2)²]^1/2

_$\phi$* _{(z, $\rho$=0)} = (M/2) {[(D/2)² + (z - l/2)²]^1/2 - [(d/2)² + (z - l/2)²]^1/2 - [(D/2)² + (z + l/2)²]^1/2 - [(d/2)² + (z + l/2)²]^1/2}

Now, using eq.1 & we get

$\vec{B}$ = - _$\mu$0 (d_$\phi$* / dt) |_z=0

$\vec{B}$ = - (_$\mu$0 M / 2) { (-l/2) / [(D/2)² + (l/2)²]^1/2 + (l/2) / [(d/2)² + (l/2)²]^1/2 - (l/2) / [(D/2)² + (l/2)²]^1/2 + (l/2) / [(d/2)² + (l/2)²]^1/2}

$\vec{B}$ = _$\mu$0 M { l / [(D² + l²)]^1/2 - l / [(d² + l²)]^1/2 }