Magnetic Vector Potential MCQ Quiz - Objective Question with Answer for Magnetic Vector Potential - Download Free PDF

We have,

$\overrightarrow{\mathrm{H}}=\frac{1}{{\mu }_0}\overrightarrow{\mathrm{\nabla }}\times \overrightarrow{\mathrm{A}}$

$=\frac{1}{{\mu }_0}\left[\begin{array}{ccc}{\hat{\mathrm{a}}}_{\mathrm{x}}& {\hat{\mathrm{a}}}_{\mathrm{y}}& {\hat{\mathrm{a}}}_{\mathrm{z}}\\ \frac{\partial }{\partial \mathrm{x}}& \frac{\partial }{\partial \mathrm{y}}& \frac{\partial }{{\partial }_{\mathrm{z}}}\\ 0& 0& -{\mu }_0\left({\mathrm{x}}^2+{\mathrm{y}}^2\right)\end{array}\right]$

$=\frac{1}{{\mu }_0}\left[-{\mu }_0\left(2\mathrm{y}\right){\hat{\mathrm{a}}}_{\mathrm{x}}-\left(-{\mu }_0\left(2\mathrm{x}\right){\hat{\mathrm{a}}}_{\mathrm{y}}\right)\right]$

$\overrightarrow{\mathrm{H}}=-2\left(\mathrm{y}{\hat{\mathrm{a}}}_{\mathrm{x}}-\mathrm{x}{\hat{\mathrm{a}}}_{\mathrm{y}}\right)\mathrm{A}/\mathrm{m}$

$\overrightarrow{\mathrm{H}}=2\mathrm{r}{\hat{\mathrm{a}}}_{\varphi }\mathrm{A}/\mathrm{m}$ where

$\mathrm{r}=\sqrt{\left({\mathrm{x}}^2+{\mathrm{y}}^2\right)}$

Now, from Ampere’s circuital law,

We have,

$\underset{\mathrm{c}}{\overset{}{\int }}\overrightarrow{\mathrm{H}}.\mathrm{d}\overrightarrow{\mathrm{l}}={\mathrm{J}}_0\pi {\mathrm{r}}^2$

$\mathrm{H}.2\pi \mathrm{r}={\mathrm{J}}_0\pi {\mathrm{r}}^2$

$\overrightarrow{\mathrm{H}}=\frac{{\mathrm{J}}_0}{2}\mathrm{r}{\hat{\mathrm{a}}}_{\varphi }\mathrm{A}/\mathrm{m}$

Comparing we have $\frac{{\mathrm{J}}_0}{2}=2$

$\underset{\mathrm{L}}{\oint }\overrightarrow{\mathrm{A}}.\mathrm{d}\overrightarrow{\mathrm{l}}=\psi =\underset{\mathrm{S}}{\int }\overrightarrow{\mathrm{B}}.\mathrm{d}\overrightarrow{\mathrm{s}}$.

Thus, the expression $\underset{\mathrm{L}}{\oint }\overrightarrow{\mathrm{A}}.\mathrm{d}\overrightarrow{\mathrm{l}},\overrightarrow{\mathrm{A}}$ is equal to the flux of magnetic flux density field across a surface bounded by loop $\mathrm{L}$.