A solenoid of length 8.0 x 10^-2 m and cross sectional area of 5.0 x 10^-5 m^2 c

Question

A solenoid of length 8.0 x 10^-2 m and cross sectional area of 5.0 x 10^-5 m^2 contains 6500 Turns per meter of length. Determine the magnitude of the emf inducted in the solenoid when the current int he solenoid changes from 0 to 1.5A during a time interval of .2 seconds.

Want to see if I'm on track here.
6500 turns per meter x 8.0 x 10^-2 meters = 520 turns
E= N (delta phi/delta t)
Binitial= (4pi x 10^-7)(520Turns/8 x10^-2m)0A= 0
Bfinal= (4pi x 10^-7)(520Turns/8 x 10^-2m) 1.5A= .012T

Phiinitial=0
Phi final= (.012T)(5 x 10^-5) = 6.1 x 10^-7

E = 520 ((6.1 x 10^-7/.2)= 1.6 x 10^-3V

Explanation / Answer

Length L = 8.0 x 10 -2 m
Cross sectional area A = 5.0 x 10 -5 m 2 Number of truns N = 6500 Change in current di = 1.5 A - 0 A                                 = 1.5 A Time interval dt = 0.2 s Inductance of the solinoid L ' = N 2 A/ L Where = Permeability of free space                = 4 x 10 -7 H / m Subsitute values we get L ' = 0.03318 H The magnitude of the emf inducted in the solenoid when the current int he solenoid changes from 0 to 1.5A during a time interval of .2 seconds E = L' di / dt                                                                  = 0.2488 V

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