Question

A fat wire, radius a, carries a constant current I, uniformly distributed over its cross section. A narrow gap in the wire, of width w << a, forms a parallel-plate capacitor. Find the magnetic field in the gap, at a distance s < a from the axis.

Answer 1

The magnetic field in the gap is $\mathbf{B = \dfrac{\mu_oI s}{2 \pi a^2}}$

In a circuit, a magnetic flux is circulated or followed via a confined environment or passage called a magnetic field gap. The narrow air gap is a non-magnetic component of a magnetic circuit that is normally connected to the remainder of the circuit magnetically in series. This enables a significant amount of magnetic flux to pass via the gap.

The magnetic field in the gap at a distance s < a can be computed by using the formula:

$\mathbf{ \oint Bdl = \mu_oI_{enclosed}}$

where;

• Magnetic flux density = B
• distance = d

$\mathbf{B( 2 \pi d) = \mu _o \oint _s J_d da }$

where;

• $\mathbf{J_d}$ = drift current density

$\mathbf{B( 2 \pi d) = \mu _o J_d \oint _sda }$

$\mathbf{B( 2 \pi d) = \mu _o J_d (\pi d^2) }$

Making the magnetic flux density the subject, we have:

$\mathbf{B =\dfrac{ \mu _o J_d (\pi d^2) }{( 2 \pi d)}}$

$\mathbf{B =\dfrac{ \mu _o J_dd}{ 2 }}$

Recall that, the drift current density $\mathbf{J_d = \dfrac{I}{\pi a^2}}$

$\mathbf{B = \dfrac{\mu_o d}{2}(\dfrac{I}{\pi a^2})}$

Recall that distance in question is said to be (s);

∴

$\mathbf{B = \dfrac{\mu_oI s}{2 \pi a^2}}$

Learn more about the magnetic field here:

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