Question

A capacitor of capacitance $0.5\text{C}$ is connected to a battery of emf $20\text{V}$ for a long time and then disconnected. The charged capacitor is then connected across a long solenoid having $1000$ turns per $\text{metre}$ in its closely packed windings on its core. After connection, it is found that the voltage across the capacitor drops to $8\text{V}$ in a time interval of $2\text{s}$. In this period, what is the average magnetic induction at the center of the solenoid.

• A. $1.8\text{mT}$
• B. $1.2\pi \text{mT}$
• C. $\pi \text{mT}$
• D. $4.5\pi \text{mT}$

Answer 1

The correct option is B $1.2\pi \text{mT}$

Initial charge on the capacitor is,

$Q_i=C{V}_i$

$Q_i=0.5\times 20=10\text{C}$

Final charge on the capacitor when the voltage across the capacitor is dropped to $8\text{V}$ is

$Q_f=0.5\times 8=4\text{C}$

$\therefore$ Charge flown through inductor in this time period will be

$Q_{\text{flown}}=Q_i-Q_f=10-4=6\text{C}$

$\therefore$ Average current in the inductor during $t=0\text{s}$ to $t=2\text{s}$ is,

$i_{avg}=\frac{Q_{\text{flown}}}{\Delta t}=\frac{6}{2}=3\text{A}$

$\therefore$ Average magnetic inductor at the centre of solenoid is

$B={\mu }_{0}n{i}_{avg}$

$B=4\pi \times {10}^{-7}\times 1000\times 3=1.2\pi \times {10}^{-3}\text{T}$

$\therefore B=1.2\pi \text{mT}$

Therefore, option $\left(\text{B}\right)$ is correct.