Question

Uniform electric flux. Figure 32-30 showsa circular region of radius R = 3.00 cmin which a uniform electric flux is directed out of the plane of the page. The total electric flux through the region is given by ${\mathbf{\varnothing }}_{\mathbf{E}}\mathbf{=}\mathbf{(}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{m}\mathbf{V}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{)}\mathit{t}$, where is tin seconds. (a)What is the magnitude of the magnetic field that is induced at a radial distance 2.00 cm?(b)What is the magnitude of the magnetic field that is induced at a radial distance 5.00 cm?

Figure 32-30

Short answer

1. The magnitude of an induced magnetic field at a radial distance 2.00 cm is$1.18\times {10}^{-19}T.$
2. The magnitude of an induced magnetic field at a radial distance 5.00 cm is$1.06\times {10}^{-19}T.$

Step-by-step solution

The given data

1. Radius of the circular region,$R=3cm\times \frac{1m}{100cm}=0.03m$
2. The total electric flux through the region,${\varphi }_E=\left(3.0\times {10}^{-3}V.m/s\right)t$
3. The direction of the total electric flux is out of the plane of the page.
4. Radial distances at which the magnetic field is induced, $r_1=2cm\times \frac{1}{100m}=0.02m$
   $r_2=5cm\times \frac{1}{100m}=0.05m$
   

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.

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

The displacement current through the wire,$i_d={\epsilon }_0\frac{d{\varphi }_E}{dt}$ (iii)

Where,${\epsilon }_0=\left(8.85\times {10}^{-12}{C}^2/N.m^2\right)$is the permittivity in vacuum,${\varphi }_E$is the electric flux through the region.

(a) Determining themagnitude of an induced magnetic field at a radial distance 2.00 cm

As the electric flux is uniform, the rate of the electric flux flowing through the region can be given as follows:

$\frac{d{\varphi }_E}{dt}=3.0\times {10}^{-3}V.m/s$

Substituting the above value in equation (iii), we can get the value of the displacement current as follows:

$\begin{array}{rcl}{i}_d& =& \left(8.85\times {10}^{-12}{C}^2/N.m^2\right)\times \left(3.00\times {10}^{-3}V.m/s\right)\\ & =& 2.66\times {10}^{-14}A\end{array}$

The radial distance r₁ = 0.02 m is situated inside the capacitor. So, using the given values substituted in equation (i), the value of the magnetic field induced can be given as follows:

role="math" localid="1663141210824" width="380" height="125">B=4π×10-7T.m/A×2.66×10-14A×0.02m2π×0.03m2=2×10-7×2.66×10-14×0.020.0009T=1.18×10-19T

Therefore, the magnitude of an induced magnetic field at a radial distance 2.00 cm is$\begin{array}{rcl}B& =& 1.18\times {10}^{-19}T\end{array}$

(b) Determining themagnitude of an induced magnetic field at a radial distance 5.00 cm

Similarly, for the r₂ = 0.05 m , it is outside the capacitor. So using the given data in equation (ii) for the magnetic field outside the capacitor can be given as follows:

$\begin{array}{rcl}B& =& \left(\frac{\left(4\mathrm{\pi }\times {10}^{-7}\mathrm{T}.\mathrm{m}/\mathrm{A}\right)\times \left(2.66\times {10}^{-14}A\right)}{2\mathrm{\pi }\times \left(0.05\mathrm{m}\right)}\right)\\ & =& \left(\frac{2\times {10}^{-7}\times 2.66\times {10}^{-14}\times 0.02}{0.05}\right)T\\ & =& 1.06\times {10}^{-19}T\end{array}$

Therefore, the magnitude of an induced magnetic field at a radial distance 5.00 cm is$\begin{array}{rcl}B& =& 1.06\times {10}^{-19}T\end{array}$