A technician wraps wire around a tube of length 38 cm having a diameter of 8.1 c

Question

A technician wraps wire around a tube of length 38 cm having a diameter of 8.1 cm. When the windings are evenly spread over thefull length of the tube, the result is a solenoid containing585 turns of wire. (a) Find the self-inductance of thissolenoid.
1 mH

(b) If the current in this solenoid increases at the rate of2 A/s, what is the self-induced emf in thesolenoid?
2 mV (a) Find the self-inductance of thissolenoid.
1 mH

(b) If the current in this solenoid increases at the rate of2 A/s, what is the self-induced emf in thesolenoid?
2 mV

Explanation / Answer

length of the loop is l = 0.38m diameter of the tube is D = 8.1cm so the radius of the tube is r = D/2 = 4.05 x10-2m cross sectional area of the tube is A = r2 calculate for A calculate for A number of turns N = 585
    0 = 4 x10-7 Tm/A
(a) The self -inductance of the loop is           L = (0N2A/l)                           = 4 x 10-7 * 585 * 585 * 3.14 * 4.05*4.05 x10-4 / 0.38
                          = 5.825 mH
(b) current in the solenoid increases at the ratedi/dt = 2 A/s        The magnitude ofself-induced emf in the solenoid is                        = L*di/dt
                         = 5.825 mH * 2A/s
                         = 11.65 mV

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