A long solenoid of length l=2.0 m , radius r=0.1 m and total number of turns N=1000 is carrying a current `i_(0)=20.0A`. The axis of the solenoid conicides with the z-axis.
(a) State the expression for the magnetic field of the solenoid and calculate its value ? Magnetic field.
(b) Obtain the expression for the self - inductance (L) of the solenoid. Calcuate its value. Value of L .
( c) Calculate the energy stored (D) when the solenoid carries this current ?
(d) Let the resistance of the solenoid be R. It is connected to a battery of emf e. Obtain the expression for the current (i) in the solenoid.
(e) Let the solenoid with resistance R describe in part (d) be strecthed at a constant speed u ( l is increased but B and `gamma` are constant ). State kirchhoff's second law for this cases. (Note: Do not solve for the current )
(f) Consider a time varying current `i=i_(0) cos (omegat)` (where `i_(0)=20.0A`) flowing in the solenoid. Obtain an expression for the electric field due to the current in the solenoid. (Note: Part(e) is not operative, i.e., the solenoid is not being stretched. )
(g) Consider `t=pi//2 omega and omega=200//pi rad-s^(-1)` in the previous part. Plot the magnitude of the electric field as a function of the radial distance from the solenoid. Also, sketch the electric lines of force.

Explanation

`N=1000,i_(0)=20A`
(a) `vecB=mu_(0)(N)/(l)i_(0)(pmhatk)`
`|vecB|=4pixx10^(-3)`Telsa
(b) `phi=Li_(0)`
BNA`=Li_(0)`
`L=(BNA)/(i_(0))=(mu_(0)N^(2)pir^(2))/(l)=2pi^(2)xx10^(-3)H`.
(C) ` E=(1)/(2)Li_(0)^(2)=(4pi^(2))/(10)` Joules=3.95J
(d) `i=i_(0)(1-e^(-t//tau))`
where `tau=(L)/(R),i_(0)=(epsi)/( R)`
(e) `epsi-iR-(d)/(dt)(Li)=0` ltbtgt `epsi-iR-L(di)/(dt)-i(dL)/(dt)`
`epsi-iR+L(di)/(dt)+i(dL)/(dt)`
after time t,
`L=(mu_(0)N^(2)pir^(2))/((l+vt))to (dL)/(dt)=-(mu_(0)N^(2)pir^(2)v)/((L+vt)^(2))`
` epsi=iR+(Ldi)/(dt)-(mu_(0)N^(2)pir^(2)v)/((L+vt)^(2))`
(f) `E_("induced")=intvecE.vec(dl)=-(dphi)/(dt)`
`intvecE.vec(dl)=-(dphi)/(dt)`
(d `to` Radial distance from axis of solenoid :
`r to ` Radius of solenoid )
For `(d lt r)`
`E.2pid=-pid^(2)(dB)/(dt)`
`=pid^(2) mu_(0)(N)/(l)(di)/(dt)`
`E.2pid=pid^(2) mu_(0) (N)/(l)i_(0)omega sin(omegat)`
`E=(mu_(0)Ni_(0)omegad sin(omegat))/(2l)`
For `d ge r `
`E. 2pid=-pi^(2)(dB)/(dt)`
`E=(mu_(0)Ni_(0)omegar^(2))/(2dl) sin(omegat)`
(g)
`t=(pi)/(2W)`
`E=(mu_(0)Ni_(0)omegad)/(2l)d lt r`
`E=(mu_(0)Ni_(0)omegar^(2))/(2dl) d ge r `
Lines of force