A long air-core solenoid of 420 turns, length 32 cm anddameter 3.0 cm carries a
ID: 1673410 • Letter: A
Question
A long air-core solenoid of 420 turns, length 32 cm anddameter 3.0 cm carries a current 3.0 A at certain time. Thecurrent is increasing at a rate of 15 A/s. a) Find the self-inductance of the solenoid b.) the voltage across the solenoid at that time c.) the magnetic energy stored in the solenoid A long air-core solenoid of 420 turns, length 32 cm anddameter 3.0 cm carries a current 3.0 A at certain time. Thecurrent is increasing at a rate of 15 A/s. a) Find the self-inductance of the solenoid b.) the voltage across the solenoid at that time c.) the magnetic energy stored in the solenoidExplanation / Answer
a ) Number of turns in a solenoid N =420 length of the solenoid l =0.32 m diameter d = 3.0 cm , radius r = 0.15cm Area A = r2 = ( 0.15 cm )2 = 0.070*10-4 m2 self inductance of asolenoid L = 0 A N2 /l = 4 *10-7 * 0.070*10-4m2 * ( 420 )2 / 0.32 m = 4.89*10-6 H b ) Voltage across the solenoid at thattime is V = L dI / dt = 4.89*10-6 H * 15 A / s = 73.3*10-6 V c ) Magnetic Energy stored in a solenoid is uB = 1/2 B2 / magnetic field in a solenoid is B =0 N i = 4 *10-7 * 420 * 3.0 A =15.8*10-4 T uB = 1/2 * ( 15.8*10-4T )2 / 4 *10-7 = 0.993 J/ m3 = 0.993 J/ m3Related Questions
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