A solenoid 24.5 cm long and with a cross-sectional area of 0.500 am2 contains 39

Question

A solenoid 24.5 cm long and with a cross-sectional area of 0.500 am2 contains 390 turns of wire and carries a current of 83.5 A. (a) Calculate the magnetic field in the solenoid. IT (b) Calculate the energy density in the magnetic field if the solenoid is filled with air. /m3 c) Calculate the total energy contained in the coil's magnetic field (assume the field is uniform) (d) Calculate the inductance of the solenoid. -14 points YF14 30 P023 An inductor with an inductance of 2.80 H and a resistance of 6.50 nis connected to the terminals of a battery with an emf 0 6.00 V and negligible internal resistance. (a) Find the initial rate of increase of current in the circuit. A/s (b) Find the raté of increase of current at the instant when the current is 0.500 A. A/s (c) Find the current 0.250 s after the dircuit is closed (d) Find the final steady-state current.

Explanation / Answer

Part A

Magnetic field inside solenoid is given by:

B = u0*n*i

n = N/L

B = 4*pi*10^-7*(390/0.245)*83.5

B = 0.167 T

Part B.

Energy density = U = B^2/(2*u0)

U = 0.167^2/(2*4*pi*10^-7)

U = 11096.68 J/m^3

Part C

Total Energy = E = U*V

V = A*L

E = 11096.68*0.5*10^-4*0.245

E = 0.136 J

Part D

E = 0.5*L*i^2

L = 2*E/i^2

L = 2*0.136/83.5^2

L = 3.90*10^-5 H

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