Class Management | Help Practice Problem 11: Inductance, LR & LC Circuits Begin

Question

Class Management | Help Practice Problem 11: Inductance, LR & LC Circuits Begin Date: 4/11 2016 12:00:00 AM - - Due Date: 12/8/2017 2:00:00 AM End Date: 12/8/2017 2:00:00 AM (10%) Problem 9: A solenoid consists of N-180 lurns of wire in a coil of length d-4 cm and cross-sectional area. 1-4 cm Assume that the magnetic lield is uniform inside the solenoid and ignore end eslects 33% Part a) Write an equation for the magnetic field B produced by the solenoid. Express the answer in terms of the current through the coil I, the number of turns N, length of the coil d, and the permeability of free space 0. rade Summary Deductions Potential 0% 100% Submissions Attempts remaining: 3 (400 per attempt) detailed view inn BACKSPACE DEL CLEAR Submit Hint give up Hints: 4% deduction per hint. Hints remaining: 3 Feedback: 5% deduction per feedback. 33% Part (b) Using the equation for the magnetic field determined in part a, calculate the magnetic flux through a slngle loop of the solenoid with a current of! 0.85 A. The coil is d = 4 cm long and has a cross-sectional area ofA = 4 cm2 and consists of N = 180 turns 33% Part (c) The self inductance relates the magnetic flux linkage to the current through the coil. Calculate the self inductance L in units of LIH. The coil is d = 4 cm long and has a cross-sectional area of = 4 cm2 and consists of N= 180 turns

Explanation / Answer

part(a)

magnetic field produced by solenoid B = uo*(N/d)*I

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part(b)

magnetic flux = B*A

magnetic flux = uo*(N/d)*I*A

magnetic flux = 4*pi*10^-7*(180/0.04)*0.85*4*10^-4

magnetic flux = 1.923*10^-6 W/m^2

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part(c)

total magnetic flux = L*I

N*1.923*10^-6 = L*I

180*1.923*10^-6 = L*0.85

self inductance L = 407.22 uH

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