Question

A capacitor of capacitance $100\mu \text{F}$ is connected to a battery of $10\text{V}$ for a long time and then disconnected from it. It is now connected across a solenoid having $200$ turns per meter. It is found that the potential difference across the capacitor drops to $90\%$ of its maximum value in $2\text{s}.$ Find the average magnetic field induced at the centre of the solenoid ?

• A. $2\pi \times {10}^{-9}\text{T}$
• B. $2\pi \times {10}^{-8}\text{T}$
• C. $4\pi \times {10}^{-9}\text{T}$
• D. $4\pi \times {10}^{-8}\text{T}$

Answer 1

The correct option is C $4\pi \times {10}^{-9}\text{T}$

Given:
$C=100\mu \text{F};V=10\text{V};n=200$
$t=2\text{s}$

Given that the capacitor is connected with the voltage source for long time, thus it will get fully charged.

Charge on the capacitor $\left(Q\right)$ is,

$Q=CV=(100\times {10}^{-6})\times 10={10}^{-3}\text{C}$

After $2\text{s}$, potential difference across the capacitor becomes $90\%$ of its maximum value.

$\therefore V^{{}^{\prime }}=\frac{90}{100}\times V=\frac{90}{100}\times 10\text{V}=9\text{V}$

Now, charge on the capacitor becomes $\left(Q^{{}^{\prime }}\right)$,

$Q^{{}^{\prime }}=C{V}^{{}^{\prime }}=(100\times {10}^{-6})\times 9=9\times {10}^{-4}\text{C}$

Therefore, current through the circuit is,

$I=\frac{Q-Q^{{}^{\prime }}}{t}=\frac{(10-9)\times {10}^{-4}}{2}=\frac{{10}^{-4}}{2}\text{A}$

$\therefore I=0.5\times {10}^{-4}\text{A}$

The magnetic field inside the solenoid is,

$B={\mu }_{0}nI$

$B=4\pi \times {10}^{-7}\times 200\times 0.5\times {10}^{-4}$

$\therefore B=4\pi \times {10}^{-9}\text{T}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.