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 0A from 4A in 0.05 s. If the resistance o

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 0A from 4A in 0.05 s. If the resistance of the coil is 10π²Ω, the total charge flowing through the coil during this time is:

Accepted Answer

32 π μC. The magnetic field inside the solenoid is given by B = μ₀nI, where n is the number of turns per unit length. The rate of change of magnetic flux through the coil is dΦ/dt = A * dB/dt, where A is the area of the coil. Using Faraday's law, the induced emf is ε = -N * dΦ/dt. The total charge is Q = ε * Δt / R. Substituting the values, Q = (100 * π * (0.01)² * μ₀ * 2 × 10⁴ * 4) / (10π²) = 32π μC.

Solution

Related NEET Physics questions