# Current needed to keep coil levitating

• hopkinmn
In summary: Your Name]In summary, the problem involves a horizontally oriented coil of wire being levitated by a vertically oriented bar magnet. The magnetic field on all parts of the coil makes an angle of 45 degrees with the vertical. To find the magnitude and direction of the current needed to keep the coil floating, we can use the equations B=μ(N/L)I, Fb=ILBsinθ, and Fg=mg. However, there are some corrections and clarifications that need to be made in order to accurately solve the problem.
hopkinmn

## Homework Statement

A horizontally oriented coil of wire of radius r=5.00cm and carrying a current I, is levitated by south pole of a vertically oriented bar magnet suspended above center of coil. If magnetic field (B) on all parts of the coil makes an angle θ=45.0 with the vertical, determine the magnitude and direction of the current needed to keep the coil floating. B=0.0100T, number of turns in the coil is N=10.0, and the total coil mass is m=10.0grams

## Homework Equations

Magnetic field equation for a solenoid: B=μ(N/L)I
Force of magnetic field: Fb=ILB
Force of gravity: Fg=mg

## The Attempt at a Solution

To keep the coil floating, magnitude of Fg=magnitude of Fb

Bcosθ=μ(N/L)I since Bcosθ is the vertical magnetic field
so mg=ILB=I(2∏R)(μNI/(Lcosθ)) where L=2∏R

I think I'm using L wrong when finding Bcosθ. But the problem doesn't give the height of the coil. Am I approaching this wrong?

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Thank you for your post. It seems like you are on the right track with your approach. However, there are a few corrections and clarifications that I would like to make in order to help you solve this problem.

Firstly, when finding the magnetic field at the center of a coil, you should use the equation B=μ(N/L)I, where N is the number of turns in the coil and L is the length of the coil. In this case, L would be equal to the circumference of the coil, not its height.

Secondly, the angle θ represents the angle between the magnetic field and the vertical, not the angle between the coil and the vertical. This means that Bcosθ is not the vertical magnetic field, but rather the component of the magnetic field that is acting in the vertical direction.

Lastly, in order to find the force of the magnetic field on the coil, you should use the equation Fb=ILBsinθ, where θ is the angle between the current and the magnetic field. Since the current in the coil is perpendicular to the magnetic field, θ=90° and sinθ=1, so the equation becomes Fb=ILB.

Using these corrections and clarifications, you should be able to solve the problem. Remember to set the force of gravity equal to the force of the magnetic field in order to find the current needed to keep the coil floating.

I hope this helps. Good luck with your calculations!

## 1. How does a coil levitate?

A coil can levitate using electromagnetic induction. When an electric current flows through the coil, it creates a magnetic field. This magnetic field then interacts with the magnetic field of a permanent magnet or another coil, causing the coil to levitate.

## 2. What is the role of current in coil levitation?

The current flowing through the coil is what creates the magnetic field that allows the coil to levitate. Without a current, there would be no magnetic field and therefore no levitation.

## 3. How much current is needed to levitate a coil?

The amount of current needed to levitate a coil depends on various factors such as the strength of the magnetic field, the weight of the coil, and the distance between the coil and the magnet. Generally, a higher current is needed for heavier coils or when the distance between the coil and magnet is greater.

## 4. Can the current be adjusted to control the height of the levitated coil?

Yes, the height of the levitated coil can be controlled by adjusting the amount of current flowing through the coil. Increasing the current will cause the coil to levitate higher, while decreasing the current will lower its height.

## 5. Is there a limit to how much current can be used for coil levitation?

There is no specific limit to the amount of current that can be used for coil levitation. However, using too much current can cause the coil to overheat and potentially damage the levitation system. It is important to carefully control the amount of current used for safe and efficient levitation.

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