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(a) At what angle $\theta$ is the torque on a current loop 90.0$\%$ of maximum? (b) 50.0$\%$ of maximum? (c) 10.0$\%$ of maximum?

Part $(a) :$ The angle is $64^{\circ}$Part $(b) :$ The angle is $30^{\circ}$Part $(c) :$ The angle is $5.7^{\circ}$

Physics 102 Electricity and Magnetism

Chapter 22

Magnetism

Magnetic Field and Magnetic Forces

Sources of Magnetic field

Electromagnetic Induction

Inductance

Cornell University

Rutgers, The State University of New Jersey

Hope College

University of Winnipeg

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So here we know that work is equaling and the number of turns and the loop times the current going through the loop a the cross sectional area of each turn in the loop times be the magnitude of the magnetic field time sign of Fada. And so, essentially, we know that the maximum torque T max occurs ATF ada equaling 90 degrees or, of course, sign of 90 degrees equaling one. And so we can simply safer part A. If the torque is 90% of its maximum, we can set sign of Fada equaling 2.9 and so fade A would simply be equaling arc sine of 0.9 and this is equaling 64.2 degrees. So when the torque is that 90% of its maximum, we know the angle is 64.2 degrees for part B. We know that the torque is only 50% of its maximum, so this is gonna be equal 2.5. And we know that from the unit circle pie here would be equaling arc sine of 0.5, which we know already to be equaling 30 degrees. And so here for part B, if the torque is only 50% of its maximum. We know that the angle is 30 degrees, and then here for Part C, it's only 10% of its maximum. So again, fate a equals arc sine of point of 0.0.1, and we find that the angle is equaling 5.74 degrees. This would be our final answer for part C. That is the end of the solution. Thank you for watching.

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