A 2.00 cm diameter 13.0 cm long solenoid has 116 turns and has a 1.00 cm diamete

Question

A 2.00 cm diameter 13.0 cm long solenoid has 116 turns and has a 1.00 cm diameter loop inside it. The loop has a resistance of 0.100 Ohm. What is the current in the loop at 0.0120 s? The maximum and the minimum current in the graph have the same magnitude which is 2.60 A. The figure provided shows a switch that has been in position 1 for a long time. If it is changed to position 2 at t = 0 s, what is the first time at which the current is maximum? L = 4.00 mH, C = 67.1 mu V = 1.11 V and R = 4.83 Ohm.

Explanation / Answer

1)
given

N = 116 turns

L = 13 cm = 0.13 m

diamter of the loop, d = 1 cm

radius of the loop, r = d/2 = 1/2 = 0.5 cm = 0.005 m

magnetic field inside a solenoid,

B = mue*N*I/L

induced emf in the loop, emf = A*dB/dt

= (pi*r^2)*(mue*N/L)*dI/dt

= (pi*0.005^2)*(4*pi*10^-7*116/0.13)*(2.6 - (-2.6))/0.02

= 2.29*10^-5 Volts

2)

frequency of Oscillation, f = 1/(2*pi*sqrt(L*C))

= 1/(2*pi*sqrt(4*10^-3*67.1*10^-6))

= 307 Hz

time period, T = 1/f

= 1/307

= 0.00326 s

time taken for the current to reach maximum, t = T/4

= 0.00326/4

= 8.15*10^-4 s

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