How to solve this coil problems?

Question

A coil is wound on an iron former of mean magnetic length of 25cm and cross-section of 4cm^2. The relative magnetic permeability is 1000, leakage, fringing and losses of all types are neglected. The coil will be connected to an AC source of frequency = 400Hz so that the peak current drawn does not exceed 0.1A and the iron operates at a peak flux density of 0.9T.

a) By using Ampere's law, determine the number of turns, N of the coil to meet this specification.

b) By using Faraday's law, obtain an expression relating the peak supply voltage to the peak magnetic field intensity.

c) Calculate the maximum permissible peak supply voltage so that the specified peak current and flux density are no exceeded.

Answer 1

a) The basic equation is H*L = I, where I is the total current enclosed. In this case I = n*i where n = no of turns, and i=current per turn (0.1A). The flux density B = µ*µ0*H, where µ0 is permeability of vacuum, and µ relative permeability of the core. Put it all together to get

n = B/(µ*µ0) * L/i = 0.9T/(4*π*10^-4 N/m^2) * 25cm / 0.1A = 1.79*10^3.

b) Faraday's law: ∫E*dl = d/dt∫B*dA
.............................line int....surface int

The first integral is just the voltage induced into each turn of the wire. The total voltage will then be n time this. B = Bp*sin(2*π*f*t), so dB/dt = Bp*2*π*f*cos(2*π*f*t). The field is uniform and covers the full area of 4cm^2, so the surface integral is 4cm^2*Bp*2*π*400*sec^-1*cos(2*π*f*t). The peak value is 4cm^2*Bp*2*π*400

V/Bp = n*4cm^2*2*π*400*sec^-1 = 1.8*10^3Volt/T

c) V= Bp*1.8*10^3Volt/T = 0.9T*1.8*10^3Volt/T = 1.62*10^3 volt

Answer 2

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