# A parallel plate capacitor with circular plates of radius $1\: m$ has a capacitance of $1\: nF$. At $t = 0$, it is connected for charging in series with a resistor $R= 1\: M\omega$ across a $2V$ battery. Calculate the magnetic field at a point $P$, halfway between the centre and the periphery of the plates, after $t = 10–3s$. (The charge on the capacitor at time $t$ is $q (t) = CV [1 – exp (–t/\tau)]$, where the time constant $\tau$ is equal to $CR$.)

$\begin {array} {1 1} (a)\;0.74 \times 10^{-13}T & \quad (b)\;7.4 \times 10^{-15}T \\ (c)\;0.74 \times 10^{-15}T & \quad (d)\;7.4 \times 10^{-13}T \end {array}$

The time constant of the $CR$ circuit is $\tau= CR= 10–3s$. Then we have
$q(t) = CV \bigg[1 – exp \bigg( \large\frac{–t}{\tau} \bigg) \bigg]$$= 2 \times 10–9 \bigg[1– exp \bigg(\large\frac{–t}{10–3} \bigg) \bigg] The electric field in between the plates at time t is E = \large\frac{q(t)}{ \in_oA}$$ =\large\frac{ q}{ \pi \in_o}$$; A = \pi (1)2 m2 = area of the plates Consider now a circular loop of radius \bigg( \large\frac{1}{2} \bigg) m parallel to the plates passing through P. The magnetic field B at all points on the loop is along the loop and of the same value.The flux \phi E through this loop is \phi E= E \times area of the loop = E \times \pi \times \bigg( \large\frac{1}{2} \bigg)2$$ = \large\frac{\pi E}{4}$$= \large\frac{q}{4\in_o} The displacement current i_d = \in_o \large\frac{d \phi E}{dt}$$ = \large\frac{1}{4}$$\large\frac{ dq}{dt}$$ = 0.5 \times 10^{-6} e^{-1} \: at \: t= 10–3s$. Now, applying Ampere-Maxwell law to the loop, we get
$B \times 2\pi \times \large\frac{1}{2}$$=\mu_o (i_c+i_d) = \mu_o(0 + i_d) = 0.5 \times 10^{-6} \mu_o e^{-1}$
$\Rightarrow B = 0.74 \times 10^{-13}T$
Ans : (a)