1. The Length of an ideal solenoid having inductance L= 4 mH consists of N=600 t

Question

1. The Length of an ideal solenoid having inductance L= 4 mH consists of N=600 turns. the cross sectional plane of the solenoid is A= 20 cm^2, and the current running thru the solenoid is 6A. What is the magnetic field (magnitude) inside of the solenoid? 2. A coil of length l=50 cm and diameter d= 5 cm consists of N= 200 turns. there is a current I= 1 A running thru the coil. Determine: a) the inductance of coil; b) the magnetic flux flowing thru a cross sectional area of the coil. 3. A coil of inductance L= 0.6H is connected with a resistor R in a simple RL circuit to a battery (DC). A switch is thrown and current starts to flow thru the circuit. After three second the current thru the coil reaches 80% of its maximum value. What is the resistance R of the RL circuit? 4. A solenoid of length 1 m and cross sectional area 20 cm^2 has an inductance of 0.4mH. Given that the magnetic energy density inside the solenoid equal 0.1J/m^3, determine the current flowing thru the solenoid.

Explanation / Answer

1. Magnetic field B inside the solenoid = µ0 * (N/l) * i l -> length of the solenoid can be calculated from the inductance(L) L= µ0 * (A/l) * N^2 therefore l = µ0 * (A/L) * N^2 therefore B = µ0 * (N/ (µ0 * (A/L) * N^2) ) * i therefore B= (i * L)/(A * N) = (6 * 4 * 10^-3)/(20 * 10^-4 * 600) = 0.02 T 2. Inductance L = µ0 * (A/l) * N^2 A = p * ( diameter^2 ) /4 µ0 = 4p×10^(-7) L= 4p×10^(-7) * ( ( p * ( diameter^2 ) /4 ) / .5) /4 =.1973 mH Magnetic field (B) inside the solenoid = n * i * µ0/l = 5.0265 * 10^-4 Flux = N * B * A = 200 * 5.0265 * 10^ -4 * p * ( diameter^2 ) /4 = 1.97392 * 10^-4

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