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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.
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Solution
Given:
Capacitance, C = 100 microfarad
Voltage, V = 20 V
Charge stored in the capacitor, Q = CV
\[= 100 \times {10}^{- 6} \times 20\]
\[ = 2 \times {10}^{- 3} C\]
It is given that the potential difference across the capacitor drops to 90% of its maximum value.
Thus,
\[V' = \frac{90}{100} \times 20 = 18 V\]
\[\text{ New charge, }Q' = CV' = 1 . 8 \times {10}^{- 3} C\]
\[\text{ Now, }\]
\[\text{ Current, i }= \frac{Q - Q'}{t}\]
\[ \Rightarrow i = \frac{(2 - 1 . 8) \times {10}^{- 3}}{2} = \frac{2 \times {10}^{- 4}}{2}\]
\[ \Rightarrow i = 1 \times {10}^{- 4} A\]
No. of turns per metre, n = 4000
Thus, the average magnetic field at the centre of the solenoid is given by
\[B = \mu_0 ni\]
\[ = 4\pi \times {10}^{- 7} \times 4000 \times {10}^{- 4} \]
\[ = 16\pi \times {10}^{- 8} \]T
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