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Question:

A longer solenoid of diameter 0.1m has 2×10⁴ turns per meter. At the center of the solenoid, a coil of 100 turns and radius 0.01m 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 0.05s. If the resistance of the coil is 10π²Ω, the total charge flowing through the coil during this time is:

32πμC

22μC

26πμC

16πμC

Solution:

ds = 0.1m, n₅ = 2×10⁴/m
nc = 100, re = 0.01m
0A from 4A in 0.05s
R = 10π²Ω
Field at center of solenoid
B = μ₀n₅i
Hence through loop Φ = BA = μ₀n₅i × π(0.01)² × nc
e = dΦ/dt = μ₀n₅ncπ(0.01)² × Δi/Δt
i = e/R, Φ = (-i) × Δt
Q = cΔt
RQ = μ₀n₅ncπ(0.01)² × Δi
Q = 4π × 10⁻⁷ × 2 × 10⁴ × 100 × π × (10⁻²)² × 4 / (10π²) × 0.05
Q = 32μC