91Ó°ÊÓ

1. The given capacitor, $C=1\mathrm{μ}\text{F or}1×{10}^{-9}\text{F}$.
2. The inductance of the coil,$L=3\text{mH or}3×{10}^{-3}\text{H}$.
3. Maximum voltage,$V_{\max }=3\text{V}$.

The maximum charge in the capacitor can be calculated by the relation between the charge, capacitor, and the potential difference across the capacitor. Also, the maximum current through the circuit is calculated by the equation for the energy of and. From the maximum current in the circuit, we can calculate the maximum energy stored in the magnetic field of the coil.

Formulae:

The charge of the capacitor,$Q=CV$ (i)

The magnetic energy stored in the inductor,$U=\frac{L{i}^2}{2}$ (ii)

The electric energy stored in the capacitor, $U=\frac{Q^2}{2C}$ (iii)

Using the given data in equation (i), we can get the maximum charge on the capacitor as follows:

$\begin{array}{rcl}{Q}_{max}& =& \left(1×{10}^{-9}\right)3\\ & =& 3×{10}^{-9}\text{C}\\ & & \end{array}$

Hence, the value of the charge is $\begin{array}{rcl}& & 3×{10}^{-9}\text{C}\end{array}$.

So, we can calculate the maximum current through the circuit by equating both the energy equations (ii) and (iii) as follows:

$\begin{array}{rcl}\frac{L{i}^2}{2}& =& \frac{Q^2}{2C}\\ i_{max}^2& =& \frac{Q_{\text{max}}^2}{LC}\\ i_{max}& =& \frac{Q_{max}}{\sqrt{LC}}\\ & =& \frac{3×{10}^{-9}}{\sqrt{3×{10}^{-3}×1×{10}^{-9}}}\\ & =& 1.73×{10}^{-3}\text{A}\end{array}$

When the current is maximum, the magnetic energy is also maximum, so using equation (ii), we can get the maximum magnetic energy in the coil can be given as follows:

$\begin{array}{rcl}{U}_{max}& =& \frac{\left(3.0×{10}^{-3}\right){\left(1.73×{10}^{-3}\right)}^2}{2}\\ & =& 4.5×{10}^{-9}\text{J}\\ & & \end{array}$

Hence, the value of the magnetic energy is $4.5×{10}^{-9}\text{J}$.