Answer to Question #140436 in Electricity and Magnetism for Kembot Lav

Question

Question: https://ibb.co/gjjB1rs
Answer: https://ibb.co/K276MzX

Answer 1

We know that the bound volume density current can be written as

$\overrightarrow{J}=\nabla \times \overrightarrow{M}$

As we know that the magnetization of bound volume current is constant hence the curl of $\overrightarrow{M}$ is zero.

As in the question, it is given that magnetization is in z axis.

$\sigma_{M}=\overrightarrow{M}.\overrightarrow{n}$

but $\hat{n}=\nabla u$

$u=\frac{x^2}{4a^2}+\frac{y^2}{4a^2}+\frac{z^2}{4b^2}-1$

$\Rightarrow \nabla u =\frac{2x}{4a^2}\hat{i} +\frac{2y}{4a^2}\hat{j}+\frac{2z}{4b^2}\hat{k}$

$\Rightarrow \nabla u =\frac{x}{2a^2}\hat{i} +\frac{y}{2a^2}\hat{j}+\frac{z}{2b^2}\hat{k}$

$\hat{n}=\frac{x}{2a^2}\hat{i} +\frac{y}{2a^2}\hat{j}+\frac{z}{2b^2}\hat{k}$

$n=\frac{\frac{x}{2a^2}\hat{i} +\frac{y}{2a^2}\hat{j}+\frac{z}{2b^2}\hat{k}}{\sqrt{(\frac{x}{2a})^2+(\frac{y}{2a})^2+(\frac{z}{2a})^2}}$

$\Rightarrow \sigma_{M}=(\overrightarrow{M_o}2b\hat{k}).\frac{\frac{x}{2a^2}\hat{i} +\frac{y}{2a^2}\hat{j}+\frac{z}{2b^2}\hat{k}}{\sqrt{(\frac{x}{2a})^2+(\frac{y}{2a})^2+(\frac{z}{2a})^2}}$

$\Rightarrow \sigma_{M}=\frac{M_oz}{b\sqrt{(\frac{x}{2a})^2+(\frac{y}{2a})^2+(\frac{z}{2a})^2}}$