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| Both sides previous revision Previous revision Next revision | Previous revision | ||
| 184_notes:examples:week12_force_loop_magnetic_field [2021/07/13 12:24] – schram45 | 184_notes:examples:week12_force_loop_magnetic_field [2021/07/13 12:33] (current) – [Solution] schram45 | ||
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| $$\vec{\tau} = IBL^2 \hat{y}$$ | $$\vec{\tau} = IBL^2 \hat{y}$$ | ||
| This sort of rotating loop is the basis for an electrical motor. Essentially you are transferring electric energy (by providing a current through the loop) to kinetic energy (by making the loop spin). | This sort of rotating loop is the basis for an electrical motor. Essentially you are transferring electric energy (by providing a current through the loop) to kinetic energy (by making the loop spin). | ||
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| + | We can check the direction of our torques by using RHR as this is also a cross product. If we put our fingers in the direction of our separation vectors and curl towards the forces, we will find that the torques are in the same direction which would cause our loop to rotate. Another way we can evaluate our solution is by looking at our forces. By now we should be very comfortable with cross products, so we should expect there to be no force on the top or bottom wire by inspection of the problem. Also, since the magnetic field is constant, both sides of the loop have the same length and the currents flow in opposite directions, we should expect the forces to be in opposite directions as well. These evaluations all line up with our solution. | ||