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Capacitance of Parallel Plate Capacitor

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Question

Show that in terms of current density $\to j$ , may be expressed as.
$- \to d B = \frac{μ_{0}}{4 π} .$ $\frac{\to j \times \to r}{r^{3}}$ $d V$

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Solution

We know that-

$\to d B = \frac{μ_{o}}{4 π} \frac{d l . \to i \times \to r}{r^{3}}$

Now, $\to i = \to j . d A$

Hence, $\to d B = \frac{μ_{o}}{4 π} \frac{\to j \times \to r}{r^{3}} d l d A$

And, $d V = d A . d l$

Thus-

$\to d B = \frac{μ_{o}}{4 π} \frac{\to j \times \to r}{r^{3}} d V$

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