91Ó°ÊÓ

a) Resistivity of the silver wire,$\mathrm{ÒÏ}=1.62×{10}^{-8}\mathrm{Î\copyright }.\text{m}$

b) Cross-sectional Area of the silver wire,

$$\begin{gathered} A=5.0{\text{mm}}^2×\frac{1{\text{m}}^2}{{10}^6{\text{mm}}^2} \\ =5.0×{10}^{-6}{\text{m}}^2 \end{gathered}$$

c) Rate of change in current in the wire,$\frac{di}{dt}=2000\text{A/s}$

d) Current of silver wire,$i=100\text{A}$.

Before charging and after charging a conductor or a capacitor, the plates will not have any magnetic field induced in the system. But when the plates are kept under a constant current flow, it induces a magnetic field in the area of the given region. This induced current is called the displacement current. The current induced consists of both real and fictitious current values. The value of the fictitious current is given by using Maxwell's equations that describe that the current magnitude is proportional to the change in the electric flux or the change in the electric field of the region.

Formulae:

The electric field due to current flowing in the wire, $E=\mathrm{ÒÏ}J$ (i)

where,$\mathrm{ÒÏ}$is the resistivity of the material,$J$is the current density of the wire.

The current density of the wire,$J=i/A$ (ii)

where,is the amount of the current flow through the wire,is the cross-sectional area of the wire.

The displacement current due to the change in the electric field,

$i_d={\mathrm{Î\mu }}_{0}A\left(\frac{dE}{dt}\right)$ (iii)

Where, ${\mathcal{E}}_0=\left(8.85×{10}^{-12}{\text{C}}^2/{\text{N.m}}^2\right)$is the permittivity in vacuum,$A$ is the cross-sectional area of the wire, $\frac{dE}{dt}$is the rate of change in the electric field.

Substituting the given values and equation (ii) in equation (i), the magnitude of the electric field can be given as follows:

$$\begin{gathered} E=\frac{\mathrm{ÒÏ}i}{A} \\ =\frac{\left(1.62×{10}^{-8}\mathrm{Î\copyright }\text{.m}\right)\left(100\text{A}\right)}{\left(5.0×{10}^{-6}{\text{m}}^2\right)} \\ =0.324\text{V/m} \end{gathered}$$

Hence, the magnitude of the electric field in wire is $E=0.324\text{V/m}$.

The value of the displacement current flowing in the wire is given using the above electric field value with the given data substituted in equation (iii) as follows:

$$\begin{gathered} i_d={\mathrm{Î\mu }}_{0}A\left(\frac{d}{dt}\frac{\mathrm{ÒÏ}i}{A}\right)\left(âˆ\mu \text{From part (a),}E=\frac{\mathrm{ÒÏ}i}{A}\right) \\ ={\mathrm{Î\mu }}_0\mathrm{ÒÏ}\left(\frac{di}{dt}\right) \\ =\left(8.85×{10}^{-12}{\text{C}}^2/{\text{N.m}}^2\right)\left(1.62×{10}^{-8}\mathrm{Î\copyright }.\text{m}\right)\left(2000\text{A/s}\right) \\ =2.87×{10}^{-16}\text{A} \end{gathered}$$

Hence, the displacement current is $i_d=2.87×{10}^{-16}\text{A}$.

The ratio of the magnitude of the magnetic field due to the displacement current to that due to the current at a distance r from the wire can be given using the values in part (b) and the given current value as follows:

$$\begin{gathered} \frac{B\left(dueto{i}_d\right)}{B\left(duetoi\right)}=\frac{i_d}{i} \\ =\frac{2.87×{10}^{-16}\text{A}}{100\text{A}} \\ =2.87×{10}^{-18} \end{gathered}$$

Therefore, the ratio of the magnetic field due to displacement current to the due to current is $2.87×{10}^{-18}$.