Question

A long solenoid of diameter 0.1 m has 2 × 10⁴ turns per meter. At the centre of the solenoid, a coil of 100 turns and radius 0.01 m is placed with its axis coinciding with the solenoid axis. The current in the solenoid reduces at a constant rate to 0 A from 4 A in 0.05 s. If the resistance of the coil is 10π² Ω, the total charge flowing through the coil during this time is

• A. 32π μC
• B. 16 μC
• C. 32 μC
• D. 16π μC

Answer 1

The Correct Option is C

To solve this problem, we need to find the total charge flowing through the coil when the current in the solenoid changes. This involves calculating the induced electromotive force (EMF) in the coil due to the changing magnetic field and then using it to determine the charge.

1. The magnetic field B inside the solenoid is given by the formula: B = \mu_0 n I where
   • \mu_0 is the permeability of free space, \mu_0 = 4\pi \times 10^{-7} \text{ Tm/A}.
   • n = 2 \times 10^4 \, \text{turns/m} is the number of turns per unit length of the solenoid.
   • I = 4 \, \text{A} is the initial current.
   Substituting the values, we have: B = (4\pi \times 10^{-7}) \times (2 \times 10^4) \times 4 = 32\pi \times 10^{-3} \, \text{T}
2. The coil is inside the solenoid, and the magnetic flux \Phi linked with the coil is given by: \Phi = B \cdot A \cdot N where
   • A = \pi r^2 is the area of the coil with r = 0.01 \, \text{m}.
   • N = 100 is the number of turns in the coil.
   So, \Phi = 32\pi \times 10^{-3} \times \pi \times (0.01)^2 \times 100 = 32\pi^2 \times 10^{-5} \, \text{Wb}
3. The change in flux when the current goes from 4 A to 0 A is the same in magnitude but opposite in direction, so the total change in flux is: \Delta \Phi = -32\pi^2 \times 10^{-5} \, \text{Wb}
4. The induced EMF (\varepsilon) in the coil is given by Faraday's law of electromagnetic induction as: \varepsilon = -\frac{\Delta \Phi}{\Delta t} where \Delta t = 0.05 \, \text{s}. So, \varepsilon = -\frac{-32\pi^2 \times 10^{-5}}{0.05} = \frac{32\pi^2 \times 10^{-5}}{0.05} = 64\pi^2 \times 10^{-4} \, \text{V}
5. Applying Ohm's Law, the total charge (Q) flowing through the coil is: Q = \frac{\varepsilon \cdot \Delta t}{R} where R = 10\pi^2 \, \Omega. Substituting the values, we have: Q = \frac{64\pi^2 \times 10^{-4} \cdot 0.05}{10\pi^2} = \frac{64 \times 10^{-4} \cdot 0.05}{10} = 32 \times 10^{-6} \, \text{C} = 32 \, \mu \text{C}

Thus, the total charge flowing through the coil during this time is 32 μC.