Q74P - Fundamentals Of Physics

Question

A parallel-plate capacitor with circular plates is being charged. Consider a circular loop centered on the central axis and located between the plates. If the loop radius of $\mathbf{\text{3.00 cm}}$ is greater than the plate radius, what is the displacement current between the plates when the magnetic field along the loop has magnitude $\mathbf{\text{2.00 μT}}$?

Step-by-Step Solution

Answer

The displacement current between the parallel plate capacitor is $i_d=\text{0.300 A}$.

Listing the given quantities:

The radius of the loop is, $r={\text{3.00 cm=3.00×10}}^{\text{-2}}\text{ m}$

The magnitude of magnetic field is, $B=\text{2.00 \hspace{0.17em}μT=}2.00\times {10}^{-6}\text{ T}$

Understanding the concepts of displacement current:

You can use the concept of the displacement current and the expression of the magnetic field outside a circular capacitor.

Formula:

$B=\frac{{\mu }_0{i}_d}{2\pi r}$

Here,  ${\mu }_0$ is the permeability of free space having a value $4\pi \times {10}^{-7} \raisebox{1ex}{$\text{T}\cdot \text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{A}$}\right.$, $i_d$ is the displacement current, and $r$ is the distance.

Calculations of the displacement current between the parallel plate capacitor:

The expression of the magnetic field outside a circular capacitor in terms of the displacement current is,

$\begin{array}{rcl}B& =& \frac{{\mu }_0{i}_d}{2\pi r}\\ i_d& =& \frac{2\pi rB}{{\mu }_0}\\ & =& \frac{2\times 3.14\times 3.00\times {10}^{-2} \text{m}\times 2.00\times {10}^{-6} \text{A}}{4\pi \times {10}^{-7} {\displaystyle \raisebox{1ex}{$\text{T}\cdot \text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{A}$}\right.}}\\ & =& 0.300 \text{A}\end{array}$

Hence, the displacement current between the parallel plate capacitor is $\begin{array}{rcl}& & 0.300 \text{A}\end{array}$.