Question

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, as shown in Fig. 7.45. Find the magnetic field in the gap, at a distance s < a from the axis.

Short answer

Answer

The magnetic field in the gap is $B=\frac{{\mu }_{0}Is}{2\pi a^2}$.

Step-by-step solution

Write the given data from the question.

The radius of the wire is a .

The constant current is the wire is I.

The gap between the wire is w<<a.

  Step 2: Determine the formulas to calculate the magnetic field in the gap. 

The expression for the current density is given as follows.

$\mathit{J}\mathbf{=}\frac{\mathbf{I}}{\mathbf{A}}$

Here, A is the area of the wire.

Calculate the magnetic field in the gap.  

Consider the figure of the wire with the gap.

Calculate the current density.

Substitute$\pi a^2$ for A into equation (1).

$J_d=\frac{I}{\pi a^2}\hat{z}$

The expression for the magnetic field from the Ampere’s law is given by,

$$\begin{gathered} \oint B·dl={\mu }_0{I}_{d_{enc}} \\ B\left(2\pi s\right)={\mu }_0{I}_{d_{enc}} \\ B=\frac{{\mu }_0{I}_{d_{enc}}}{2\pi s} \end{gathered}$$

Substitute $J_d\left(\pi s^2\right)$ for $I_{d_{enc}}$into above equation.

$$\begin{gathered} B=\frac{{\mu }_0{J}_d\left(\pi s^2\right)}{2\pi s} \\ B=\frac{{\mu }_0{J}_{d}s}{2} \end{gathered}$$

Substitute$\frac{I}{\pi a^2}$for$J_d$ into above equation.

$$\begin{gathered} B=\frac{{\mu }_0\frac{I}{\pi a^2}s}{2} \\ B=\frac{{\mu }_{0}Is}{2\pi a^2} \end{gathered}$$

Hence the magnetic field in the gap is$B=\frac{{\mu }_{0}Is}{2\pi a^2}$.