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Problem

22.11 More Applications of Magnetism Indicate wh…

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Averell H.
Carnegie Mellon University

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Problem 1 Problem 2 Problem 3 Problem 4 Problem 5 Problem 6 Problem 7 Problem 8 Problem 9 Problem 10 Problem 11 Problem 12 Problem 13 Problem 14 Problem 15 Problem 16 Problem 17 Problem 18 Problem 19 Problem 20 Problem 21 Problem 22 Problem 23 Problem 24 Problem 25 Problem 26 Problem 27 Problem 28 Problem 29 Problem 30 Problem 31 Problem 32 Problem 33 Problem 34 Problem 35 Problem 36 Problem 37 Problem 38 Problem 39 Problem 40 Problem 41 Problem 42 Problem 43 Problem 44 Problem 45 Problem 46 Problem 47 Problem 48 Problem 49 Problem 50 Problem 51 Problem 52 Problem 53 Problem 54 Problem 55 Problem 56 Problem 57 Problem 58 Problem 59 Problem 60 Problem 61 Problem 62 Problem 63 Problem 64 Problem 65 Problem 66 Problem 67 Problem 68 Problem 69 Problem 70 Problem 71 Problem 72 Problem 73 Problem 74 Problem 75 Problem 76 Problem 77 Problem 78 Problem 79 Problem 80 Problem 81 Problem 82 Problem 83 Problem 84 Problem 85 Problem 86 Problem 87 Problem 88 Problem 89 Problem 90 Problem 91

Problem 57 Easy Difficulty

Figure 22.64
Find the direction and magnitude of the force that each
wire experiences in Figure $22.64(\mathrm{b}),$ using vector addition.

Answer

top left:
$$
F_{1}=11.56^{-5} \mathrm{N} / \mathrm{m}
$$
at an angle of
$$
\alpha_{1}=26.56^{\circ}
$$
to the up of the right side of the horizontal
top right:
$$
F_{2}=11.56^{-5} \mathrm{N} / \mathrm{m}
$$
at an angle of
$$
\alpha_{2}=10.9^{\circ}
$$
to the up of the left side of the vertical
bottom right:
$$
F_{4}=5.59^{-5} \mathrm{N} / \mathrm{m}
$$
at an angle of
$$
\alpha_{3}=63.43^{\circ}
$$
to the down of the left side of the horizontal
bottom left:
$$
F_{4}=5.59^{-5} \mathrm{N} / \mathrm{m}
$$
at an angle of
$$
\alpha_{4}=63.43^{\circ}
$$
to the down of the right side of the horizontal

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Problem 15
Problem 16
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Problem 30
Problem 31
Problem 32
Problem 33
Problem 34
Problem 35
Problem 36
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Problem 45
Problem 46
Problem 47
Problem 48
Problem 49
Problem 50
Problem 51
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Problem 53
Problem 54
Problem 55
Problem 56
Problem 57
Problem 58
Problem 59
Problem 60
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Problem 63
Problem 64
Problem 65
Problem 66
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Problem 75
Problem 76
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Problem 79
Problem 80
Problem 81
Problem 82
Problem 83
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Problem 85
Problem 86
Problem 87
Problem 88
Problem 89
Problem 90
Problem 91

Video Transcript

here. We know that this would be the diagram of the system. And we know that the magnetic field is equaling the mag native permeability in a vacuum multiplied by the current divided by two pi times are are being the shortest distance to that wire. This would be the magnetic field produced by a long straight wire. And so from this diagram, we can first say that the distance D is going to be equaling to the square root of 0.10 meters quantity squared plus again 0.10 meters quantity squared. And this distance is equally one. Rather, we can say 14 points. Actually, you simply say 0.1414 meters. Now we know that here four part fort Rather, we could say the magnetic permeability sub one. This would be according to eye sub one. Now, you can actually say here that this would be equaling two. The same thing that we're using this equation here and we're essentially going to say for pie times 10 to the negative seventh test the meters per amps. This would be multiplied by here. 10 amps divided by two pi multiplied by our 0.1414 meters and we find that beast of one is equaling 1.4 14 times 10 to the negative fifth. Tesla's Keep track of all these values. Of course. Here, we're going to then say that be sub three. In this case, this is the magnetic field at oh due to wire C. And so we can say that then this would be equaling four pi times 10 to the negative seventh Tesla's meters per amps multiplied by five amps. Again, this is do the wire C. This would be divided by two pi multiplied by the shortest distance, again being 20.1414 meters. And we find that the magnetic field at oh due to wire C is equaling 7.7 times, 10 to the negative sixth. It might be easier to write 0.707 times 10 to the negative fit test list and from symmetry, we know that these up to is equaling beasts of one where be sub four is equal and be sub three. And so the magnetic field on the X axis would be equaling piece of one plus piece of two plus piece of three plus piece of four multiplied by co sign of 45 degrees. And we can say that to be sub X is gonna be equaling 1.414 times, 10 to the negative fifth plus 1.414 times, 10 to the negative fifth plus 0.707 times 10 to the negative fifth plus 0.707 times, 10 to the negative fifth. This would, of course, be for Tesla's and then Times Co sign of 45 degrees. And we find that the magnetic field on the X axis is equaling 3.0 times, 10 to the negative fifth test floods and the Y direction. By symmetry, this would be equaling. Be sub one minus b sub two minus B said three plus be sub four. And you know that this would be times sign 45 degrees. But however you know that all these you're gonna cancel out and this would equal zero Tesla's so here we can simply say that the magnitude of the magnetic field is equaling 3.0 times, 10 to negative fifth Tesla's and, of course, Fada with the equaling zero degrees because this entire magnitude is in the ex direction. This would be our final answer. That is the end of the solution. Thank you for one.

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Irina Lyublinskaya, Gregg Wolfe, Douglas Ingram , Liza Pujji

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