Question

Find the value of magnetic field between the plates of a capacitor at distance $1m$ from the center where, the electric field varies by ${{10}^{10}}V{{m}^{-1}}$ per second.
(A). $5.56\times {{10}^{-8}}T$
(B). $5.56\times {{10}^{-3}}T$
(C). $5.56\mu T$
(D). $5.56T$

Answer 1

The electric field in a parallel plate capacitor varies as ${{10}^{10}}V{{m}^{-1}}$. The magnetic field inside the capacitor depends on absolute permittivity, absolute permeability and rate of change of magnetic field. Substituting corresponding values in the relation, we can calculate the magnetic field. Also the product of permittivity and permeability is the reciprocal of square of speed of light in air.

Formulas used:
${{c}^{2}}=\dfrac{1}{{{\mu }_{0}}{{\varepsilon }_{0}}}$
$B=\dfrac{{{\mu }_{0}}{{\varepsilon }_{0}}}{2}\dfrac{dE}{dt}$

Complete step by step answer:
The electric field in the capacitor is changing by ${{10}^{10}}V{{m}^{-1}}$ per second. Therefore,
$\dfrac{dE}{dt}={{10}^{10}}V{{m}^{-1}}$
We have to calculate magnetic field at a point $1m$ from the center
We know that,
${{c}^{2}}=\dfrac{1}{{{\mu }_{0}}{{\varepsilon }_{0}}}$
Here, $c$ is the speed of light in air
${{\mu }_{0}}$ is the permeability of free space
${{\varepsilon }_{0}}$ is the permittivity of free space
The permeability of a medium is the measure of how much it gets magnetized when kept in an external magnetic field.
The permittivity of a medium is the measure of electrical energy stored in the presence of an electric field.
We know that, $c=3\times {{10}^{8}}m{{s}^{-1}}$
Substituting in above equation, we get,
$\begin{aligned} & {{(3\times {{10}^{8}})}^{2}}=\dfrac{1}{{{\mu }_{0}}{{\varepsilon }_{0}}} \\\ & \Rightarrow {{\mu }_{0}}{{\varepsilon }_{0}}=\dfrac{1}{9\times {{10}^{16}}} \\\ \end{aligned}$
Between the plates of a capacitor, the magnetic field is calculated as-
$B=\dfrac{{{\mu }_{0}}{{\varepsilon }_{0}}}{2}\dfrac{dE}{dt}$
Substituting given values in the above equation, we get,
$\begin{aligned} & B=\dfrac{{{\mu }_{0}}{{\varepsilon }_{0}}}{2}\dfrac{dE}{dt} \\\ & \Rightarrow B=\dfrac{1}{9\times {{10}^{16}}\times 2}{{10}^{10}} \\\ & \Rightarrow B=0.0556\times {{10}^{-6}}T \\\ & \therefore B=5.56\times {{10}^{-8}}T \\\ \end{aligned}$
Therefore, the magnetic field between the plates of capacitors is $5.56\times {{10}^{-8}}T$.
Hence, the correct option is (A).

Note: A parallel plate capacitor consists of two metals plates with a dielectric or free space between them and the plates are oppositely charged. The ability of a capacitor to store charge inside it is known as the capacitance. The capacitance of a parallel plate capacitor depends on its dimensions and the permittivity of the medium inside it.