Question: Figure 32-35a shows the current i that is produced in a wire of resistivity$\mathbf{1}\mathbf{.}\mathbf{62}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathit{\Omega }\mathit{m}$ . The magnitude of the current versus time i_s = 10.0A is shown in Figure b. The vertical axis scale is set by t_s= 50.0ms and the horizontal axis scale is set by . Point P is at radial distance 9.00 mm from the wire’s centre. (a) Determine the magnitude of the magnetic field ${\overrightarrow{\mathbf{B}}}_{\mathbf{i}}$at point P due to the actual current i in the wire at t= 20 ms, (b) At t = 40 ms, and (c) t = 60 ms . Next, assume that the electric field driving the current is confined to the wire. Then Determine the magnitude of the magnetic field at point P due to the displacement current i_d in the wire at (d) t = 20 ms , (e) At t = 40 ms , and (f) At t= 60 ms . At point P at 20 s, what is the direction (into or out of the page)of (g) ${\overrightarrow{\mathbf{B}}}_{\mathbf{i}}$and (h) ${\overrightarrow{\mathbf{B}}}_{\mathbf{id}}$?

1. Resistivity of the given wire, $\rho =1.62\times {10}^{-8}\text{.m}$
2. Current value in the vertical scale of the graph,$i_s=10\text{A}$
3. Time value in the horizontal scale of the graph,$t_s=50\text{ms}$
4. Radial distance of the point P from the center of the wire, $r=9\text{mm}\times \frac{1\text{m}}{{10}^3\text{mm}}=0.009\text{m}$

A current-carrying wire induces a magnetic field within a radial distance at a point due to its property that the charges are always moving within the conductor. Using Lenz law and experiment it can be found that the current flow induced flow such that they oppose the magnetic field induced. The direction of the induced magnetic field due to the current-carrying wire is found to be perpendicular to the direction of the current. This can be justified using the right-hand thumb rule. If the thumb is in the direction of the current and the fingers point to the point at which the magnetic field is induced, then the direction in which the fingers are curled will give the magnetic field due to the current wire.

Formulae:

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

$i_d={\epsilon }_{0}A\left(\frac{dE}{dt}\right)$ (i)

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

The electric field due to current flowing in the wire, $E=\rho J$ (ii)

where, is the resistivity of the material, is the current density of the wire.

The magnetic field at a radial point due to a current carrying wire,

$B=\frac{{\mu }_{0}i}{2\pi r}$ (iii)

where, ${\mu }_0=\left(4\pi \times {10}^{-7}\text{T.m/A}\right)$is the magnetic permittivity constant, is the current flowing in the wire, $J=i/A$ is the radial distance of the point from the center axis of the wire.

The current density of the wire, (iv)

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

From the given graph, it can be seen that for the horizontal time value , ,t = 20 ms the current value is in the vertical scale i = 4A.

Thus, using the above data in equation (iii), the value of the magnetic field at point P can be given as follows:

$$\begin{gathered} B_i=\frac{\left(4\pi \times {10}^{-7}\text{T.m/A}\right)\left(4\text{A}\right)}{2\pi \left(9\times {10}^{-3}\text{m}\right)} \\ =8.9\times {10}^{-5}\text{T} \\ =0.089\text{mT} \end{gathered}$$

Therefore, the magnitude of the magnetic field ${\overrightarrow{\mathrm{B}}}_{\mathrm{id}}$ at point P due to the actual current i in the wire at $t=20ms,is0.089\text{mT}.$

From the given graph, it can be seen that for the horizontal time value t = 40 ms , the current value is in the vertical scale i = 8A .

Thus, using the above data in equation (iii), the value of the magnetic field at point P can be given as follows:

$$\begin{gathered} B_i=\frac{\left(4\pi \times {10}^{-7}\text{T.m/A}\right)\left(8\text{A}\right)}{2\pi \left(9\times {10}^{-3}\text{m}\right)} \\ =0.18\text{mT} \end{gathered}$$

Therefore, the magnitude of the magnetic field ${\overrightarrow{\mathrm{B}}}_i$ at point P due to the actual current i in the wire at t = 40 ms is $0.18\text{mT}$

When the time value in horizontal axis is t > 50 ms , the current is constant.

Thus, From the given graph, it can be seen that for the horizontal time value t = 60 ms , the current value is in the vertical scale i = 10 A.

Thus, using the above data in equation (iii), the value of the magnetic field at point P can be given as follows:

$$\begin{gathered} B_i=\frac{\left(4\pi \times {10}^{-7}\text{T.m/A}\right)\left(10\text{A}\right)}{2\pi \left(9\times {10}^{-3}\text{m}\right)} \\ =0.220\text{mT} \end{gathered}$$

Therefore, the magnitude of the magnetic field ${\overrightarrow{\mathrm{B}}}_{\mathrm{id}}$ at point P due to the actual current i in the wire at, t = 60 ms is $0.220\text{mT}$

Using equation (iv) in equation (i), the value of the electric field within the region can be given as follows:

$$\begin{gathered} \rho =\frac{AE}{i} \\ E=\frac{\rho i}{A} \end{gathered}$$

Thus, the displacement current within the region using the above value in equation (i) becomes,

$i_d={\epsilon }_0\rho \left(\frac{di}{dt}\right)$

From the graph, the rate of change of current can be given as:

$$\begin{gathered} \frac{di}{dt}=\frac{\text{2A}}{\text{10ms}} \\ =\text{200 A/s} \end{gathered}$$

This is only for the region $0\text{s}<t<50\text{s}$

Thus, the displacement current using this value becomes:

$$\begin{gathered} i_d=\left(8.85\times {10}^{-12}{\text{C}}^2{\text{/N.m}}^2\right)\left(1.62\times {10}^{-8}\Omega m\right)\left(\text{200}\frac{\text{A}}{\text{s}}\right) \\ ==2.9\times {10}^{-17}\text{A} \end{gathered}$$

Now, using the above data in equation (iii), the value of the magnetic field at point P can be given as follows:

$$\begin{gathered} B_{id}=\frac{\left(4\pi \times {10}^{-t}\right)\left(2.9\times {10}^{17}A\right)}{4\pi \left(9\times {10}^{-3}\right)} \\ =6.4\times {10}^{\text{- 22}}\text{T} \end{gathered}$$

Therefore, the magnitude of the magnetic field ${\overrightarrow{\mathrm{B}}}_{\mathrm{id}}$ at point P due to the displacement current i in the wire at, t = 20 ms is $6.4\times {10}^{-22}\text{T}.$

As for the region$0\text{s}<t<50\text{s}$ , the value of the current change is constant, that is given by the calculations in part (d) as:

$\frac{di}{dt}=\text{200 A/s}$

Thus, would be the same as calculated in (d).

$B_{id}=6.4\times {10}^{-22}\text{T}$

Therefore, the magnitude of the magnetic field at point P due to the displacement current i_d in the wire at, t = 40 ms is $6.4\times {10}^{-22}\text{T}.$

In this region, there is no rate of change of current, i.e.$\frac{di}{dt}=0\text{A/s}$,

Now, using the above data in equation (iii), the value of the magnetic field at point P can be given as follows:

$B_{id}=0\text{T}$
Therefore, the magnitude of the magnetic field ${\overrightarrow{B}}_{id}$ at point P due to the displacement current i_d in the wire at, t = 60 s is 0T

It can seen that the current is in rightward direction and thus pointing the current in the thumb direction such that the fingers are in the direction pointing to the point and then curling then will given the field direction to be coming out of the page.

Using the right-hand curl rule, the direction of magnetic field $\overrightarrow{{\mathrm{B}}_{\mathrm{i}}}$ at t = 20 ms is out of the page.

Therefore, at point P at t = 20 s , direction of the magnetic field of, $\overrightarrow{{\mathrm{B}}_{\mathrm{i}}}$ is out of the page.

It can seen that the current is in rightward direction and thus pointing the current in the thumb direction such that the fingers are in the direction pointing to the point and then curling then will given the field direction to be coming out of the page.

Thus, using the right-hand rule, the direction of magnetic field ${\overrightarrow{B}}_{id}$ at t = 20 ms is out of the page.

Therefore, at point P at t = 20 s , direction of the magnetic field of, ${\overrightarrow{B}}_{id}$ is out of the page.