Physics 1b - Thermal Physics & Electromagnetism

H. E. Smith |
Spring 2000 |

Physics 1B - Tutorial #10 |

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**The figure above shows two coils, A & B. Use Lenz' Law to determine which direction current flows through the resistor when**- (a) the switch is closed.
- (b) the switch is opened after being closed.
- (c) the current in A decreases.
- (d) the current in A increases.
**(e) B is moved toward A.**In (a), (d), (e) the flux of magnetic field toward the right is increasing, by Lenz' Law B will act to oppose the increasing flux, by creating a flux/field in the opposite direction - current will be right to left.

In (b) & (c) current will be left to right.

- The conducting loop shown above has area A = 1m
^{2}(1m on a side) and is moving with v=1m/s into a magnetic field B = 10^{-4}T. Graph the current in the loop as a function of time.

From Faraday's Law

= -d/dt = B^{.}dA/dt = Blv = 10^{-4}T^{.}1m^{.}1m/s = 10^{-4}V

as the loop enters the field. If*emf*and current are positive counter-clockwise, the current will behave as shown on the left, with I_{1}= 10^{-4}A. **An apartment complex is supplied by single-phase AC from a power plant 5-km away. The wire that runs from the power plant to the complex and back has a resistance per unit length of 1 x 10**^{-5}/m, supplying 100,000W.**Calculate the resistive power losses in the wire if the AC is supplied at 110V.**The total wire resistance is 1 x 10^{-5}^{.}(5000m + 5000m) = 0.1 . The power*supplied*is P = VI so that I = 100,000W/110V = 910A. The power lost to heating the wire due to its resistance is P = I^{2}R = (910A)^{2}^{.}0.1 = 83kW. In other words, the power plant has to produce 183kW in order to deliver 100kW, losing 45% to resistive heating.**Calculate the resistive power losses in the wire if the AC is supplied at 500kV (typical of US electric distribution).**In this case the current is much lower, I = 100,000W/500,000V = 0.2A and the resistive losses, P = (0.2A)^{2}^{.}0.1 = 0.004W, are negligible.**Design a way to transform AC provided at 500kV to 110V.**You need a*transformer*, one of those big cylinders that you see mounted on power poles which uses magnetic induction such that V_{source}/V_{household}= N_{s}/N_{h}, where N is the number of turns of wire on the appropriate side of the transformer. In this case, V_{source}/V_{household}= 500,000V/110V = 4545, so the*turn ratio*of this*step-down transformer*is N_{s}/N_{h}= 4545 (*e.g.*if the output - 110V - side of the transformer has 100 turns, the input side would need 454,545 turns).

Gene Smith

Last modified: Tues., 6 June 2000

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Tutorial #9**** **

Gene Smith

Last modified: Tues., 7 June 2000