Calculate efficiency I know the input power V⋅I = 36 W supplied to e-magnet but how to get the mechanical power required to keep the object attached up to e-magnet?
Alisa Taylor
Answered question
2022-10-20
I energize it so that 1lb steel object gets attached to it. I energize it with DC current from a battery so that object only barely kept attached but does not press against the e-magnet, in other words the power provided to e-magnet produces only equal to the gravitational force of the object, no more. Assuming that the vectors of forces are fully vertical, force developed by e-magnet is equal to the weight of the object.
To calculate efficiency I know the input power W supplied to e-magnet but how to get the mechanical power required to keep the object attached up to e-magnet?
Answer & Explanation
Kason Gonzales
Beginner2022-10-21Added 15 answers
In your magnet setup, it is not the electrical power that is holding up the steel object, but the magnetic field created by the electrical current. If you used wire with a lower electrical resistance, less power would be required to maintain the same current. In fact, if you used superconducting wire, then no power would be needed since the current would persist without need of a battery. This is actually done for electromagnets in the Large Hadron Collider. Without electrical resistance, no power is lost, so no power input is required.
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