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QPages 249-253: Q61: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Due to a Current Pages 249-253: Q61 Solution: A capacitor of capacitance 100 �F is connected to a battery of…

⚠️ Common Exam Traps & Conceptual Pitfalls

Do not attempt to solve the complex differential equation for the $LC$ circuit to find the instantaneous current; instead, use the definition of average current $I_{avg} = \Delta Q / \Delta t$ derived from the change in capacitor charge over the given time interval.

A capacitor of capacitance 100 �F 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.

Step-by-Step Verified Solution

**Answer:**

B = 1.1 \times 10^{-5}

T. **Step-Step Explanation:**

C = 100 \mu

F,

V = 20

V,

N = 4000

turns/m. Initial charge

Q_0 = CV = 2 \times 10^{-3}

C. After 2 s,

V' = 0.9V = 18

V. Charge

Q = 1.8 \times 10^{-3}

C. Average current

i = \frac{\Delta Q}{\Delta t} = \frac{2 \times 10^{-4}}{2} = 10^{-4}

A.

B = \mu_0 n i = 4\pi \times 10^{-7} \times 4000 \times 10^{-4} = 5 \times 10^{-7}

T.

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