A 42.0 mA current is carried by a uniformly wound air-core solenoid with 490 tur

Question

A 42.0 mA current is carried by a uniformly wound air-core solenoid with 490 turns, a 13.0 mm diameter, and 12.0 cm length, (a) Compute the magnetic field inside the solenoid. (b) Compute the magnetic flux through each turn. T-m^2 (c) Compute the inductance of the solenoid. mH (d) Which of these quantities depends on the current? (Select all that apply.) magnetic field inside the solenoid magnetic flux through each turn inductance of the solenoid A 485-turn solenoid has a radius of 7.90 mm and an overall length of 13.8 cm. (a) What is its inductance? mH (b) If the solenoid is connected in series with a 2.50-ohm resistor and a battery, what is the time constant of the circuit?

Explanation / Answer

3)I =42 mA, N =490 , d =13 mm , r = 6.5 mm , l = 12 cm

(A) B = uoNI/l = (4*3.14*10^-7*490*0.042)/(0.12)

B =215 uT

(b) flux = B.A = 215*10^-6*3.14*6.5*6.5*10^-6

Flux = 2.85*10^-8 T.m^2

(c) magnetic filed inside the solenoid

magnetic flux through each turn

4) N =485 , r =7.9 mm, l =13.8 cm

(a) L =uoN^2A/l

L = 4*3.14*10^-7*485*485*3.14*7.9*7.9*10^-6/(0.138)

L =0.42 mH

(b) R =2.5 ohm

Time constant T = L/R = 0.00042/2.5

T= 0.168 ms

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