Homework9_Solutions.pdf - Homework 9 Solutions ECE 333 Professor Amir Arbabi Problem 1 L \u2248 \u00b50 D 2 \u0001\u0002 ln 8D 2 \u0001 \u0003 \u22122 B = 10\u22123 cos(2\u03c0f t)\u02c6 z(T(a

Homework9_Solutions.pdf - Homework 9 Solutions ECE 333...

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Solutions Homework 9 ECE 333 Professor Amir Arbabi Problem 1 L μ 0 ( D 2 ) ln ( 8 D 2 ) - 2 B = 10 - 3 cos(2 πft ) ˆ z (T) (a) Use Faraday’s Law and draw schematic of a circuit describing the voltage current relation in the loop First, looking at Faraday’s law we note that I E · d l = - d Φ dt (1) = - d dt Z S B · d s (2) = - d dt Z S ( B ext + B ind ) · d s (3) = - d Φ ext dt - d Φ ind dt (4) Where the subscript ext refers to the external magnetic field imposed on our loop, and the subscript ind refers to the induced effects the loop exerts in response to the field B By recognizing that the contour integral of E will be zero along the wire and nonzero at the load, R We can rewrite the result as follows: 1
Solutions Homework 9 ECE 333 Professor Amir Arbabi I E · d l + d Φ ind dt = - d Φ ext dt Where H E · d l is the voltage across the resistor R , d Φ ind dt = IL is voltage change due to self-inductance, and - d Φ ext dt is our sinusoidal voltage source. This resembles an equation for KVL analysis for a lumped element network, and we can indeed sketch a circuit diagram to represent this equation (below) - d Φ ext dt + - L + - 250Ω Where L can be approximated by: L μ 0 D 2 ln 8 D d - 2 = (4 π × 10 - 7 ) . 15 2 ln 8( . 15) . 001 - 2 = 4 . 80 × 10 - 7 and our source - d Φ ext dt can be found via: - d Φ ext dt = - d dt Z S B · d s = - d dt Z S B · ˆ z rdrdφ = - d dt Z 2 π 0 Z . 15 0 10 - 3 cos(2 πft ) ˆ z · ˆ z rdrdφ = 2 πf 10 - 3 sin(2 πft ) Z 2 π 0 Z . 15 0 rdrdφ = 4 . 5 π 2 f × 10 - 5 sin(2 πft ) 2
Solutions Homework 9 ECE 333

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