5. A coil of Nc turns and radius b surrounds a long solenoid...

5. A coil of Nc turns and radius b surrounds a long solenoid of radius a and n turns per meter in Figure. The current in the solenoid changes as I0 = 10e^sin(wt). Circuit has a variable resistor. (a) Find the induced emf (electromotor force) in the coil as a function of time. (b) Find the self-inductance of solenoid. (c) Find the potential V of the circuit when at t=2 s resistor has 2 ?, L=0.4 H, and f =10 Hz. (At t=2 s variable resistor can be used like a normal resistor) Solenoid, n turns per meter I0 = 10e^sin(wt) 2 ? Coil, Nc turns I0 = 10e^sin(wt) Variable resistor

Transcript

00:01 This is a question based on solenoid where we have the magnetic field due to primary winding is given by b which is equal to mu0 into n into i0.

00:11 So, this is for primary windings.

00:15 So, primary winding or primary coil or primary coil then we have the magnetic field due to or the flux due to secondary coil is given by it is b into nc into ap that is the area of the primary coil.

00:33 So, we have this is due to so it is the flux in the secondary coil.

00:39 So, on substituting the values this is mu0 into n into i0 into nc into pi into a square where a is the radius of the primary cell.

00:51 Then we have change in emf will be by faraday's law we can write it as minus l into d pi by dt where l is the inductance.

00:59 So, here we have this is minus l into d by dt of mu0 n i0 nc pi into a square.

01:08 So, here all terms are constant except current therefore this is minus l into mu0 n nc pi into a square d i0 by dt.

01:21 Then we have this can be written as it is minus l into mu0 n nc pi into a square.

01:29 So, i0 is given as it is so it is d by dt of 10 e power sin omega t.

01:35 So, 10 is constant sin omega t.

01:38 So, e power so this is exponential of so this is the value of exponential of sin omega t into differentiation of sin omega t is cos omega t into omega.

01:51 Then we have the value of inductance as minus 10.

01:54 So, l into mu0 or here taking this n taking this 10 with the other terms a square e power sin omega t into cos omega t into omega.

02:13 Then we have the formula for the self -inductance is given by it is pi divided by i0.

02:19 So, it is pi is d n nc into pi a square divided by ia...

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