A long solenoid with n turns per meter and radius R has a time-dependent current

Question

A long solenoid with n turns per meter and radius R has a time-dependent current flowing through its coil. The cross section of the solenoid is shown above. Obtain an expression for the electric field as a function of r inside the solenoid. Make a sketch of the cross sectional view of the solenoid and draw the electric field lines in the solenoid. The solenoid has 10 turns per centimeter, a radius of R=5.0 mm, and is driven by a current at a frequency of f=60 Hz with peak current I0=1.00 A. What is the peak value of the electric field at a radius of 1.0 mm inside the solenoid?

Explanation / Answer

B = magnetic field inside solenoid

= mu*n* I

= mu*n* Io* cos(wt)

A = area of loop of radius r = pi* r^2

Flux = flux through loop

       = B*A

       = (pi*r^2) * mu*n* Io* cos(wt)

emf = induced voltage in loop

       = -d(flux)/dt

       = (pi*r^2) * mu*n* Io*w* sin(wt)

a) E = eletric filed

     integration[ E* dL] = emf

             E * (2*pi* r)   = emf

             E = emf/(2*pi*r)

             E   = [ (pi*r^2) * mu*n* Io*w* sin(wt) ] /(2*pi*r)

              E = r*mu*n* Io*w* sin(wt)/2

b ) electrid field is along the tangent to the loop

c)   E(peak) = peak value of electric field

                   = r*mu*n* Io*w*/2

           given r = 1 mm= 10^-3 m

, n = 10 turn /cm = 10 *10^2 turn/m , Io = 1 A ,    W = 2*pi* f = 2*pi* 60 = 120 pi rad/s

hence   E(peak) =( 10^-3 )*(4*pi*10^-7) * ( 10 *10^2)*1*(120 pi)/2

                         = 2.36*10^-4 N/C

         

                 

     

   

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