Question

A capacitor of capacitance 100 µF is connected to a battery of 20 volts for a long time and then disconnected from it. It is now connected across a long solenoid having 4000 turns per metre. It is found that the potential difference across the capacitor drops to 90% of its maximum value in 2.0 seconds. Estimate the average magnetic field produced at the centre of the solenoid during this period.

Answer 1

Explanation:

Step 1:

Given data in the question  

Capacitance of battery C = 100$\mu F$

Voltage of a battery  V = 20 V

Charge held in the condenser,Q = CV

$=100 \times 10^{-6} \times 20$

$=2 \times 10^{-3} \mathrm{C}$

Step 2:

The potential difference across the condenser is given to drop to 90 per cent of its maximum value.

$V^{\prime}=\frac{90}{100} \times 20=18 V$

$Q^{\prime}=C V^{\prime}=1.8 \times 10^{-3} \mathrm{C}$

$Q^{\prime}$ is new charge of the battery  

$i=\frac{Q-Q^{\prime}}{t}$

 $i=\frac{(2-1.8) \times 10^{-3}}{2}=\frac{2 \times 10^{-4}}{2}$

$i=1 \times 10^{-4} A$

Number of turns per metre, n = 4000

Step 3:

Consequently, the average magnetic field at the solenoid centre is given by

$B=\mu_{0} n i$

$=4 \pi \times 10^{-7} \times 4000 \times 10^{-4}$

$=16 \pi \times 10^{-8} T$