Q7.34P - Introduction to Electrodynamics

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.

Step-by-Step Solution

Answer

Answer

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

Step 1: 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.

Step 3: 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$B=\frac{{\mu }_{0}Is}{2\pi a^2}$ .