GATE 2026 PH – Question 61
A capacitor is made of two circular metal plates of radius 1 m separated by a distance of $d=1$ mm. The space between them is filled by a dielectric with permittivity $\epsilon_r=5$. The capacitor is connected to a voltage $V=10\sin(2\pi\times10^6t)$ volts, where $t$ is in seconds. The maximum value of the magnetic field in between the plates at a radial distance $r=0.5$ m from the centre of the capacitor is $B\times10^{-6}$ T. The value of $B$ (rounded off to two decimal places) is _____ (Speed of light in vacuum $c=3\times10^8\ \mathrm{m\,s^{-1}}$)
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Correct answer: 0.86 to 0.88
Explanation
Between the capacitor plates there is a changing electric field. Inside the plates, the displacement current acts as a source of a circulating magnetic field. Use the **Ampère-Maxwell law** over a circle of radius $r$ (symmetry):
$$B\,(2\pi r)=\mu_0\,\epsilon_0\epsilon_r\,\frac{dE}{dt}\,(\pi r^2)\;\Rightarrow\;B=\frac{\mu_0\epsilon_0\epsilon_r\,r}{2}\,\frac{dE}{dt}=\frac{\epsilon_r\,r}{2c^2}\frac{dE}{dt}.$$
**Electric field:** $E=\dfrac Vd=\dfrac{10\sin(\omega t)}{d}$ with $\omega=2\pi\times10^6$ rad/s, so the maximum value of $dE/dt$ is
$$\left|\frac{dE}{dt}\right|_{max}=\frac{10\,\omega}{d}=\frac{10\times2\pi\times10^6}{10^{-3}}=6.283\times10^{10}\text{ V m}^{-1}\text{s}^{-1}.$$
**Maximum magnetic field** at $r=0.5$ m, with $\epsilon_r=5$ and $c^2=9\times10^{16}$ m²/s²:
$$B_{max}=\frac{5\times0.5}{2\times9\times10^{16}}\times6.283\times10^{10}=\frac{2.5\times6.283\times10^{10}}{1.8\times10^{17}}=8.73\times10^{-7}\text{ T}.$$
So $B\times10^{-6}$ T with $B=\mathbf{0.87}$.