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September 13, 2025, 02:30:38 am

Author Topic: AOS 1 Question - important !!!!!!  (Read 492 times)  Share 

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FlorianK

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AOS 1 Question - important !!!!!!
« on: August 07, 2012, 11:10:03 pm »
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We got our SAC back and there are 2 questions, which are basically the same. Where my teacher is saying I got them wrong. He wasn't able to explain to me why they'd be wrong, which is why I still think I'm right.

Question 1:
Calculate the maximum EMF generated by a coil of 250 turns and cross-sectional area 30cm², rotating inside a magnetic field of strength 0.45 T, if it takes 1.5 milliseconds (ms) for one rotation.

My solution: (Including my thoughts)
Average Voltage for 90° turn: (90° and dividing the time by 4, because that is what it needs to go from maximum to mimimum flux)
(250*0.003m²*0.45T)/(0.0015s/4)=900V
Peak Voltage: (Times sqrt(2) because the question is asking for maximum EMF, therefor peak voltage)
900V*sqrt(2)=1272.8V

What I actually wrote was:
(250*0.003m²*0.45T*4*sqrt(2))/(0.0015s)=1272.8V
My teacher circeld the "4*sqrt(2)" and wrote next to it "what is this doing there".

My teacher "model solution":
(250*0.003m²*0.45T)/0.0015s=225V

Question 2:
A coil has dimensions 0.3m*0.4m. It has 5 turns. It is in a uniform magnetic field of 5.0 T. The coil is rotated at a constant rate of 10 revolutions per second.
What is the average voltage developed across the resistor R when the coil rotates through 90° from orientation shown above? With orientation shown above is meant the diagram, which showed maximum flux.

My solution: (Including my thoughts)
E=N delta(flux)/delta(time)
E=5*(0.3*0.4*5)/(0.1/4)=120V - 'divided by 4' like in question 1

My teacher completly wrong solution "model solution":
E=N delta(flux)/delta(time)
E=E=5*(0.3*0.4*5)/(0.1)=30V
30V=1 rotation
7.5V=1/4 rotation

Especially his solution for question 2 has to be wrong, since it wouldn't be 60V for 2 rotations.

thank you
Florian