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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 $CR$.)


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


1 Answer

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)
answered Mar 4, 2014 by thanvigandhi_1

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