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course PHY 232
4/28 3:50 pm I didn't realize I was this far behind on my submissions. I'm going to focus on the laboratory questions more than the calculations or answers clearly given in the notes.
110309 Physics II`q005. When you crank the generator at some constant rate:
Is it harder to crank when the clips are together or apart?
**** Together
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Is it harder to crank when current is flowing or when current is not flowing?
****Flowing
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Would it make sense to say that the resistance of the circuit to the electrical current is high, or when it is low?
**** Low because more energy is used to move more charges and R makes charge flow less meaning less energy needed to move the charge
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Is harder cranking therefore associated with high or low electrical resistance?
**** Lower resistance
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`q006. According to what you observed, does a compass directly below a current tend to line up with the current, or perpendicular to it?
****It was perp
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`q007. Do you think you are able to distinguish, just by cranking the generator, whether the clips are far apart or close together on the metal bracket? How could you test this without biasing the experiment by perceiving what you expect to perceive?
`q008. When the battery created a current through just the wire lead, did the compass react more or less than when the current included the generator?
**** I couldn't distinguish very much but seemed easier the farther apart they are. Better: use battery and Ammeter
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... compare with result when generator handle held stationary
University Physics only:
`q009. What was the `r vector from the domino to your nose?
**** r = <1, 0.5, 0.3>
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What therefore would be the magnetic field produced by the 'domino segment' have been at the tip of your nose, had the 3 amp current in fact been flowing through the bar?
**** B ≈ k (I `d L x u_r/ r^2)
B = 5.64x10^-8 Tesla
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Suppose the bar extended to infinity in both directions, and you were to analyze the contributions of 'domino segments' lined up all along the bar. How would you approach the problem?
****Form sum of `d B's and integrate
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`q010. Suppose we arrange a number of 'domino segments' to form a circle whose radius is equal to the length of your previous `r vector, with the circle in a horizontal plane centered at the tip of your nose. Consider the current to still be 3 amps, and assume that it is moving counterclockwise around the circle.
What would be the magnitude and direction of the magnetic field due to one of these segments, at the center of the circle (i.e., at the tip of your nose)?
****The direction is up since from rt hand rule
Break into x and y components and use x only since in same plane and integrate around the circumference
B = k I*2πr^2/(r^2)^3/2
B = 1.63x10^-5 Tesla
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How many segments would there be in the entire circle, and what therefore would be the magnitude and direction of the total magnetic field at the center of the circle?
**** Oops did most of this above but 7.2 segments
@& Dominoes are only 5 cm long, so there would be more than 7.2 of them around the circle.
However you clearly understand this.*@
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`q011. Now consider the magnetic field at the point on the floor directly below the center of the circle. Use a reasonable estimate of the distance from the tip of your nose to the floor.
Consider first the field due to the domino in which the the current is in the `i direction. What is the `r vector from that domino to the point on the floor, and what therefore are the magnitude and direction of the magnetic field at that point, due to that domino?
****∆B = k (I `d L x u_r)/r^2
r= <0, 0, -1.25> meters
@& The domino is on the circle, which is some distance away from your nose. The vector from your nose to the point is mostly parallel to the xy plane. So the vector from the domino to the point directly beneath your nose will have a significant component parallel to the x-y plane. The vector you give here has zero component parallel to the xy plane.*@
radius = 1.16 m
r_f= <1.6, 0, -1.25> m
`d B = k* (I `d L x u_f)/r_f^2
`d B = 6.72*10^-28 T
Direction is still up due to CCW I flow
@& The r vector from domino to the point on the floor has a significant component parallel to the xy plane, and a significant z component as well. So its cross product with the current vector will have significant components parallel to the xy plane (as a result of the z component or the r vector) and in the z dinrection (as a result of the component of the z vector parallel to the xy plane).*@
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Now answer the same for the domino on the opposite side of the circle (the one you get to if you move along a line from the original domino through the center of the circle, continuing to the other side of the circle)?
**** Same in mag and direction
@& The z component will be the same as befpre. but the component in the xy plane will be opposite to that obtained before.
So the z components add, the components in the xy plane cancel.*@
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What is the vector sum of the two magnetic fields?
**** <0, 1.34*10^-28, 0>
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Can you extend this reasoning to reason out the magnitude and direction of the magnetic field at the given point, due to the entire circular loop?
****Direction is up bc of rt hand and magnitude is only the components in the z direction
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`q012. How much force would that domino segment experience in a magnetic field of .1 Tesla perpendicular to its length?
**** T = N/amp*m --> amps = 3, m = 0.15 --> F = 0.045 Ns
@& The calculation would be Tesla * amps * meters, which would have units of Newtons.*@
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If the field was directed vertically downward, what would be the direction of the force, assuming the domino to be oriented north-south and the current to be flowing to the south?
**** East
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`q013. The original domino created a small magnetic field at the position of your notes. If there was another domino at the position of your nose, orientated in the north-south direction, with the current running to the south, how much force would it experience as a result of that field, and what would be the direction of that force?
**** It would depend on the angle the two made with each other where the closer to perp the less interaction
@& Assuming the orgiinal domino to be oriented north-south, the field at the second domino would be either up or down, perpendicular to the current in that domino.*@
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How would your answer change if the current was running toward the north?
**** Just in the opposite direction as above
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What if it was running east to west?
**** Again it would only effect components in similar direction and according to the angles formed.
&#Good work. See my notes and let me know if you have questions. &#