A long solenoid of radius 2 cm has 100 turns/cm and is surrounded by a 100 turn coil of radius 4 cm having a total resistance 20Ω. If current changes from 5 A to − 5A, find the charge through galvanometer.
Question
A long solenoid of radius 2 cm has 100 turns/cm and is surrounded by a 100 turn coil of radius 4 cm having a total resistance 20Ω. If current changes from 5 A to − 5A, find the charge through galvanometer.
Solution
Sure, here is the step by step solution:
Step 1: First, we need to calculate the magnetic field inside the solenoid. The formula for the magnetic field inside a solenoid is B = μ0 * n * I, where μ0 is the permeability of free space, n is the number of turns per unit length, and I is the current.
Step 2: The change in magnetic field ΔB is then given by ΔB = B(final) - B(initial) = μ0 * n * I(final) - μ0 * n * I(initial). Given that the current changes from 5 A to -5 A, ΔB = μ0 * n * (-5 A) - μ0 * n * (5 A) = -2 * μ0 * n * 5 A.
Step 3: The change in magnetic flux ΔΦ through the coil is given by ΔΦ = ΔB * A, where A is the area of the coil. The area of the coil is given by A = π * r^2, where r is the radius of the coil. So, ΔΦ = -2 * μ0 * n * 5 A * π * (0.04 m)^2.
Step 4: The electromotive force (emf) induced in the coil is given by Faraday's law of electromagnetic induction, which states that the emf is equal to the rate of change of magnetic flux. So, emf = - ΔΦ / Δt, where Δt is the time it takes for the current to change.
Step 5: The current I in the coil is then given by Ohm's law, I = emf / R, where R is the resistance of the coil.
Step 6: Finally, the charge Q through the galvanometer is given by Q = I * Δt.
Note: The actual numerical solution will depend on the value of μ0 (the permeability of free space), and the time Δt it takes for the current to change, which are not given in the problem.
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