Q19P - Fundamentals Of Physics

Question

The figure 32-30 shows a circular region of radius $\mathbf{R}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{cm}$  in which a displacement current is directed out of the page. The displacement current has a uniform density of magnitude  (a) What is the magnitude of the magnetic field due to displacement current at a radial distance  2.00 cm ?(b) What is the magnitude of the magnetic field due to displacement current at a radial distance  5.00 cm?

                 

Step-by-Step Solution

Answer

a) The magnitude of the magnetic field due to displacement current at a radial distance $R=2.00\text{ cm}$ at is $B=75.4\text{ nT}$.

b) The magnitude of the magnetic field due to displacement current at a radial distance $R=5.00 \text{cm}$ at  is $B = 67.9 nT$

The given data

a) Displacement current density,  $J_d=6.00 A/m^2$

b) The radius of the circular region,

  $$\begin{gathered} R=3.00 cm \times \frac{1 m}{100 cm} \\ =3.00\times {10}^{-2}m \end{gathered}$$

c) Radial distances at which the magnetic field is induced,   

$$\begin{gathered} r=2cm\times \frac{1}{100 m} \\ =0.02 m \end{gathered}$$

$$\begin{gathered} r_2=5\text{ cm}\times \frac{1}{100\text{ m}} \\ =0.05\text{ m} \end{gathered}$$

Understanding the concept of induced magnetic field

When a conductor is placed in a region of changing magnetic field, it induces a displacement current that starts flowing through it as it causes the case of an electric field produced in the conductor region. According to Lenz law, the current flows through the conductor such that it opposes the change in magnetic flux through the area enclosed by the loop or the conductor. The magnitude of the magnetic field is due to the displacement current using the displacement current density when the Amperian loop is smaller and larger than the given circular area.

Formulae:

The magnetic field at a point inside the capacitor,   $B=\left(\frac{{\mu }_0{i}_{d}r}{2\pi R^2}\right)$                 (i)

where,  is the magnetic field, ${\mu }_0=\left(4\pi \times {10}^{-7}\text{ T.m/A}\right)$ is the magnetic permittivity constant, $r$ is the radial distance, $i_d$ is the displacement current, $R$ is the radius of the circular region.

The magnetic field at a point outside the capacitor,  $B=\left(\frac{{\mu }_0{i}_d}{2\pi r}\right)$                  (ii)

Where, 

${\mu }_0=\left(4\pi \times {10}^{-7}\text{ T.m/A}\right)$ Is the magnetic permittivity constant, $r$ is the radial distance,  and $i_d$ is the displacement current.

The current flowing in a given region,   $i = JA$                                             (iii)

Where, 

 J Is the current density of the material, $A$ is the cross-sectional area of the material.

(a) Determining the magnitude of the magnetic field due to displacement current at a radial distance

The area of a circular plate can be given as follows:

 $A=\pi R^2$

Thus, the value of the displacement current can be given using the above data in the equation (iii) as follows:

 $i_d=J_d\pi R^2$

For the given radial distance , $r_1=0.02\text{ m}$  , $r_1<R$the magnitude of the magnetic field can be given using the above current value and the given data substituted in equation (i) as follows:

 $$\begin{gathered} B=\frac{{\mu }_0{J}_d\pi R^2{r}_1 }{2\pi R^2} \\ =\frac{{\mu }_0{J}_d{r}_1 }{2} \\ =\frac{\left(4\pi \times {10}^{-7}\text{ T.m/A}\right)\times \left(6.00{\text{ A/m}}^2\right)\times \left(0.02\text{ m}\right) }{2} \\ =7.54\times {10}^{-8}\text{ T} \\ =75.4\text{ nT} \end{gathered}$$

Therefore, the magnitude of the magnetic field due to displacement current at a radial distance  $R=2.00 cm$is $B=75.4 \text{nT}$.

(b) determining the magnitude of the magnetic field due to displacement current at radial distance

For the given radial distance , $r_2=0.05\text{ m}$ , $r_2>R$,the magnitude of the magnetic field can be given using the above current value from part (a) and the given data substituted in equation  (ii) as follows:

 $$\begin{gathered} B=\frac{{\mu }_0{J}_d\pi R^2 }{2\pi r_2} \\ =\frac{{\mu }_0{J}_d{R}^2 }{2r_2} \\ =\frac{\left(4\pi \times {10}^{-7}\text{ T.m/A}\right)\times \left(6.00{\text{ A/m}}^2\right)\times \left({0.03}^2{\text{ m}}^2 \right)}{2\times \left(0.05\text{ m}\right)} \\ =6.79\times {10}^{-8}\text{ T} \\ =67.9\text{ nT} \end{gathered}$$

Therefore, the magnitude of the magnetic field due to displacement current at a radial distance  $R=5.00 cm$is $B=67.9 \text{nT}$