Question

Find the value of magnetic field between the plates of a capacitor at a distance \(1\,m\) from centre, where electric field varies by \(10^{10}\,V/m\) per second.

• A. \(5.56\times 10^{-8}\,T\)
• B. \(5.56\times 10^{-9}\,T\)
• C. \(5.56\,\mu T\)
• D. \(5.55\,T\)

Hint:

Magnetic field due to displacement current: \(B=\dfrac{\mu_0\varepsilon_0 r}{2}\dfrac{dE}{dt}\). Use \(\mu_0\varepsilon_0=\dfrac{1}{c^2}\).

Answer 1

The Correct Option is A

Step 1: Use Maxwell’s displacement current concept.
Between capacitor plates, magnetic field is due to displacement current.
Using Ampere-Maxwell law:
\[ B(2\pi r) = \mu_0\varepsilon_0 \frac{d\Phi_E}{dt} \] Step 2: Electric flux through area.
\[ \Phi_E = EA = E(\pi r^2) \Rightarrow \frac{d\Phi_E}{dt} = \pi r^2 \frac{dE}{dt} \] Step 3: Substitute in equation.
\[ B(2\pi r) = \mu_0\varepsilon_0 (\pi r^2)\frac{dE}{dt} \Rightarrow B = \frac{\mu_0\varepsilon_0 r}{2}\frac{dE}{dt} \] Step 4: Insert values.
\[ \mu_0\varepsilon_0 = \frac{1}{c^2} = \frac{1}{(3\times 10^8)^2} = \frac{1}{9\times 10^{16}} \] \(r=1\,m\), \(\dfrac{dE}{dt}=10^{10}\).
\[ B = \frac{1}{2}\cdot\frac{1}{9\times 10^{16}}\cdot 1\cdot 10^{10} = \frac{10^{10}}{18\times 10^{16}} = \frac{1}{18}\times 10^{-6} \approx 5.56\times 10^{-8}\,T \] Final Answer: \[ \boxed{5.56\times 10^{-8}\,T} \]