91Ó°ÊÓ

The radius of a circle, $R=3\text{cm}$

The magnitude of displacement current, $i_d=\left(3\text{A}\right)\left(\frac{r}{R}\right)$

Determine the displacement current for the non-uniform current density using Ampere’s -Maxwell law. Evaluate the magnitude of the magnetic field using the displacement current. For the uniform displacement current density, determine the magnitude of the magnetic field for the circular region of radius R for the cases: By using the relation between the current density and the displacement current, and the expression of flux in terms of area and electric field, find the current density as a function of the electric field. In electromagnetism, displacement current density is the quantity appearing in Maxwell equations defined in terms of the rate of change of D, the electric displacement field.

1. $\mathit{r}\mathbf{<}\mathit{R}$
2. $\mathit{r}\mathbf{>}\mathit{R}$

where $\mathit{r}$is the radius of the Amperian loop.

Formulae are as follows:

$âˆ\circledR \stackrel{→}{B}.d\stackrel{→}{S}={\mathrm{μ}}_0{i}_{d,enc}+{\mathrm{μ}}_0{i}_{enc}$

where, $\stackrel{→}{B}$ is the magnetic field.

The Ampere-Maxwell law is given by,

$âˆ\circledR \stackrel{→}{B}.d\stackrel{→}{S}={\mathrm{μ}}_0{i}_{d,enc}+{\mathrm{μ}}_0{i}_{enc}$

The magnitude of the displacement current is given, and the second term of the above equation is zero. Therefore,

$âˆ\circledR \stackrel{→}{B}.d\stackrel{→}{S}={\mathrm{μ}}_0{i}_{d,enc}$

In this case, to find the magnetic field inside $\left(r<R\right)$the circular region of the radius $r$.

So draw the Gaussian region inside the circle of radius $R$.

The magnetic field at this circle is constant, so take $B$outside the integral.

$Bâˆ\circledR d\stackrel{→}{S}={\mathrm{μ}}_0{i}_{d,enc}$

$B\left(2\mathrm{π}r\right)={\mathrm{μ}}_0{i}_{d,enc}$

$B=\frac{{\mathrm{μ}}_0{i}_{d,enc}}{2\mathrm{π}r}$

Substituting $i_{d,enc}=\left(3\text{A}\right)\left(\frac{r}{R}\right)$

$B=\frac{{\mathrm{μ}}_0\left(3\text{A}\right)\left(\frac{r}{R}\right)}{2\mathrm{π}r}$

$B=\frac{{\mathrm{μ}}_0\left(3\text{A}\right)}{2\mathrm{π}R}$

Substituting the values,

$$\begin{gathered} B=\frac{(4\mathrm{Ï€}×{10}^7\mathrm{Tm}IA)\left(3A\right)}{22\mathrm{Ï€}(0.0300\mathrm{m})} \\ B=2×{10}^{-5}T \\ B=20\mathrm{μ°Õ} \end{gathered}$$

Therefore, the magnitude of the magnetic field due to displacement current at a radial distance $R\text{= 2.00cm}$is $B=20\mathrm{μ°Õ}$.

The magnitude of a magnetic field due to displacement current at a radial distance $5\mathrm{cm}$:

In this case, the Gaussian region is outside the circular region R and the current enclosed is the displacement current.

So the magnitude of the magnetic field at a point outsideis

$$\begin{gathered} B=\frac{{\mathrm{μ}}_0{i}_d}{2\mathrm{Ï€}r} \\ B=\frac{\left(4\mathrm{Ï€}×{10}^{-7}\text{T.m/A}\right)\left(3\text{A}\right)}{2\mathrm{Ï€}\left(0.0500\text{m}\right)} \\ B=12\mathrm{μ°Õ} \end{gathered}$$

Therefore, the magnitude of the magnetic field due to displacement current at a radial distance $R=5\mathrm{cm}$is $B=12\mathrm{μ°Õ}$.