Exam 2 practice problems solutions

Exam 2 practice problems solutions - Exam 2 practice...

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Exam 2 practice problems solutions Physics 204 1. A current flows upward in a long straight wire. To the right is a rectangular loop that is also carrying a current (the power supply is not shown so the diagram doesn’t get too cluttered). Predict the magnetic force the straight wire exerts on the rectangular loop. I need to break the loop into 4 straight segments. The magnetic force on the top segment will point up. The magnetic force on the bottom segment will point down and is equal in magnitude to the magnetic force on the top segment, so they cancel. The force on the left segment points to the left, and the force on the right segment points to the right. These do not cancel because the magnetic field is stronger closer to the straight wire. I’ll call the current in the straight wire wire I and the current in the loop loop I .        ,, 00 0 7 5 sin90 22 11 2 4 10 5.00 14.0 0.20 2 0.026 0.10 7.96 10 B B L B R loop L loop R wire wire B loop L R loop LR loop wire B B B F F F I LB I LB II F I L B B I L rr I I L F Tm A A A m F mm FN                    The minus sign indicates the force is to the left. 2. Two parallel wires are carrying currents as shown in the diagram. The electrons that make up the current 2 I move at a very slow average speed, in this case approximately 5 6.9 10 ms . a. Draw a picture of the magnetic field generated by 1 I . b. What is the magnitude and direction of the magnetic force, if any, that is exerted on one of the moving electrons in the wire on the right. The magnetic field at the position of wire 2 points into the page and has a magnitude of 01 2 I B r The magnetic force an electron in wire 2 will feel will be towards wire 1. You can use the right hand rule to figure that out, but remember that the electron is moving in the opposite direction of the current and is negatively charged. The magnitude of this magnetic force is              7 19 5 29 sin sin 90 2 4 10 2.00 1.6 10 6.9 10 1 2 0.1 4.42 10 B B B B F q vB I F ev r T m A A F C m s m   10.0 cm 1 2.00 IA 2 3.00
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That may seem ridiculously small, but the electron’s mass is about 100 times smaller than this, so this would cause the electron to accelerate at roughly 2 100 ms 3. Imagine that a space mission has been planned to send a probe to the planet Mercury. One of the goals of the mission is to measure the magnetic field of the planet as a function of altitude above the north magnetic pole of the planet. To do this, the space probe is being designed with a ring made of 1000 turns of conducting wire. This ring has a radius of 2.00 m and a total resistance of 10.0 . As the probe descends this ring will be kept parallel to the surface of the planet. You estimate that the probe will take 300 seconds to descend from orbit (where the magnetic
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Exam 2 practice problems solutions - Exam 2 practice...

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