A long solenoid has n = 400 turns per meter and carries a current given by I = 3

Question

A long solenoid has n = 400 turns per meter and carries a current given by I = 30(1-e^-1.6t) where I is in amperes and t is in seconds. Inside the solenoid and coaxial with it is a coil that has a radius of R = 6cm and consists of a total of N = 250 turns of fine wire. A) What emf is induced in the coil by the changing current? B) Write an expression for the inductance per unit length for this solenoid. C)What back emf per unit length is generated in the solenoid by the time varying current as mentioned above?

Explanation / Answer

A) E = NdFLUX/dT = 250* 4*3.14*10^-7*400*3.14*(6/100)*(6/100)*dI/dt

=> E= 250* 4*3.14*10^-7*400*3.14*(6/100)*(6/100)*30*1.6*e^-(1.6t) = 0.0681495552*e^-(1.6t) ANSWER

B) E = Ldi/dt

=> L = E/di/dt = 250* 4*3.14*10^-7*400*3.14*(6/100)*(6/100) = 0.0014197824 H/m ANSWER

C) back emf = Ldi/dt = 0.0681495552*e^-(1.6t) ANSWER

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