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Problem

Repeat Exercise $22.41,$ but with the loop lying …

06:52
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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 48 Medium Difficulty

(a) $\mathrm{A} 200$ -turn circular loop of radius 50.0 $\mathrm{cm}$ is vertical, with its axis on an east-west line. A current of 100 A circulates clockwise in the loop when viewed from the east. The Earth's field here is due north, parallel to the ground, with a strength of $3.00 \times 10^{-5} \mathrm{T} .$ What are the direction and magnitude of the torque on the loop? (b) Does this device have any
practical applications as a motor?

Answer

Part $(\mathrm{a}) :$ The torque on the loop is 0.471 $\mathrm{N}$ . $\mathrm{m}$
Part $(\mathrm{b}) :$ The torque is very small so the loop will not have any practical applications as a motor.

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Physics 102 Electricity and Magnetism

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Chapter 22

Magnetism

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Video Transcript

So here the torque would be equaling and the number of turns and the loop times I the current through the loop times air a the cross sectional area of one turn times be the magnitude of the magnetic field times sign of data. We know that here a in this case is equaling pi r squared. This would be the cross sectional area of a circular loop. And so we can say that the torque would then be equal number of loops, which was 200 the current of 100 amps times the area of pie times the radius which we know to be 50 centimeters or 500.50 meters quantity squared, multiplied by the magnetic field of three 0.0 times, 10 to the negative. So here the torque would be equaling and the number of turns and the loop times I the current through the loop times air a the cross sectional area of one turn times be the magnitude of the magnetic field times sign of data. We know that here a in this case is equal in pi r squared. This would be the cross sectional area. Tesla's multiplied by sine of 90 degrees. We know this to be one. And so the torque is going to be equaling 0.471 Newton meters downwards. This would be our answer for part a four part B. Does this application Does this device have a practical application? Yes, of a circular loop. And so we can say that the torque would then be equal number of loops, which was 200 the current of 100 amps times the area of pie times the radius which we know to be 50 centimeters or 500.50 meters quantity squared, multiplied by the magnetic field of three 0.0 times 10 to the negative. This device has a practical application, has a motor essentially, if Tesla's multiplied by sine of 90 degrees, we know this to be one, and so the torque is going to be equaling 0.471 Newton meters downwards. This would be our answer for part a four part B. Does this application Does this device have a practical application? Yes, the loop is connected to a wire than as the current passes through all the loops. The magnetic field exerts of torque on these loops which in turn, rotates the shaft. So hear this device can convert electrical energy to mechanical work. Which mechanic work being that it actually rotates the shaft? That is the end of the solution. Thank you for watching.

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