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A potential difference varying at the rate of $6 \times {10^2}{\text{V}}{{\text{s}}^{ - 1}}$ is applied to the plates of a conductor of capacity \[2\mu F\]. The magnetic field at the edge of the plate in the gap if the radius of plate is $10cm$ is
(A) \[2.4 \times {10^{ - 9}}T\]
(B) $3 \times {10^{ - 12}}T$
(C) $3.33 \times {10^{ - 9}}T$
(D) \[2.4 \times {10^{ - 6}}T\]

seo-qna
Last updated date: 18th Jun 2024
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Answer
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Hint: To solve this question, we need to find out the displacement current because of the changing potential difference between the plates of the capacitor. Then using the formula for the magnetic field due to an infinitely long current carrying wire, we can get the final answer.

Formula used: The formula used to solve this question is given by
$C = \dfrac{{{\varepsilon _0}A}}{d}$, here $C$ is the capacitance of a parallel plate capacitor having plates of area $A$ separated by a distance of $d$.

Complete step-by-step solution:
We know that the capacitance of a parallel plate capacitor is given by
$C = \dfrac{{{\varepsilon _0}A}}{d}$............(1)
We know that the area of a circular plate is given by
$A = \pi {r^2}$
Putting this in (1) we get
$C = \dfrac{{{\varepsilon _0}\pi {r^2}}}{d}$
$ \Rightarrow d = \dfrac{{{\varepsilon _0}\pi {r^2}}}{C}$ ………...(2)
Now, we know that
$E = \dfrac{V}{d}$
Differentiating both sides with respect to time t, we have
$\dfrac{{dE}}{{dt}} = \dfrac{1}{d} \times \dfrac{{dV}}{{dt}}$
Substituting (2) in the above equation we get
$\dfrac{{dE}}{{dt}} = \dfrac{C}{{{\varepsilon _0}\pi {r^2}}}\dfrac{{dV}}{{dt}}$...............(3)
Now, we know that the displacement current is given by
${i_d} = {\varepsilon _0}\dfrac{{d{\varphi _E}}}{{dt}}$...............(4)
The electric flux is given by
${\varphi _E} = EA$
$ \Rightarrow {\varphi _E} = \pi {r^2}E$
Putting this in (4) we have
${i_d} = {\varepsilon _0}\dfrac{{d\left( {\pi {r^2}E} \right)}}{{dt}}$
$ \Rightarrow {i_d} = {\varepsilon _0}\pi {r^2}\dfrac{{dE}}{{dt}}$........(5)
Now, the magnetic field at the edge of the plate can be given by
$B = \dfrac{{{\mu _0}{i_d}}}{{2\pi r}}$
Putting (5) in the above expression we get
$B = \dfrac{{{\mu _0}{\varepsilon _0}\pi {r^2}}}{{2\pi r}}\dfrac{{dE}}{{dt}}$
$ \Rightarrow B = \dfrac{{{\mu _0}{\varepsilon _0}r}}{2}\dfrac{{dE}}{{dt}}$
Putting (3) in the above expression we get
$B = \dfrac{{{\mu _0}{\varepsilon _0}r}}{2}\dfrac{C}{{{\varepsilon _0}\pi {r^2}}}\dfrac{{dV}}{{dt}}$
$ \Rightarrow B = \dfrac{{{\mu _0}C}}{{2r}}\dfrac{{dV}}{{dt}}$.............(6)
Now, according to the question, we have the radius of the plate equal to $10cm$. So we have
$r = 10cm$
$ \Rightarrow r = 0.1m$ …………..(7)
Also, the capacity of the capacitor is given to be equal to \[2\mu F\]. So we have
$C = 2\mu F$
$ \Rightarrow C = 2 \times {10^{ - 6}}F$..........(8)
Also the variation of voltage with time is given to be equal to $6 \times {10^2}{\text{V}}{{\text{s}}^{ - 1}}$. So we have
$\dfrac{{dV}}{{dt}} = 6 \times {10^2}{\text{V}}{{\text{s}}^{ - 1}}$............(9)
Substituting (7), (8), and (9) in (7), we get the final value of the magnetic field as
$B = 3.33 \times {10^{ - 9}}T$

Hence, the correct answer is option C.

Note: The magnetic field obtained from the displacement current is similar to that of the magnetic field produced by an infinitely long current carrying wire. The displacement current passes through the centre of the plates of the capacitor, and hence we obtained the magnetic field at the edge of the plate.