The heat dissipated in the wire of a solenoid.

Your name]In summary, we are asked to show that the heat dissipated in the windings of a long solenoid is independent of the size of the wire used, given a fixed magnetic field density along the axis of the solenoid. Using the equations for power, magnetic field, and resistance of a wire, we can show that the cross-sectional area of the wire does not affect the power dissipated, thus proving our statement.
  • #1
deusexlumina
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Homework Statement


The turns for a long solenoid are wound to fill the space between two concentric cylinders of fixed radii, r1 and r2. Show that, for a given magnetic field density along the axis of the solenoid, the heat dissipated in the windings is independent of the size of the wire used. For simplicity, assume wire of rectangular cross-section, that the thickness of any insulation can be neglected and that (r2 - r1) << r1.

Homework Equations



Electric field due to a solenoid
power due to current flowing through wire
Resistance of a wire as a function of length and cross sectional area

The Attempt at a Solution



Used P = I^2 * R

Re-arranged equation for B field of a solenoid so it was in terms of current. Substituted this equation and resistance of a wire equation into equation for power. Cant seem to make much headway after this though. any help would be appreciated.
 
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  • #2


Hello,

Thank you for your question. I would be happy to help you with this problem.

First, let's start by defining some variables. Let's call the number of turns in the solenoid N, the current flowing through the solenoid I, the radius of the inner cylinder r1, the radius of the outer cylinder r2, and the length of the solenoid l.

Next, let's look at the equation for the magnetic field inside a solenoid. We know that the magnetic field inside a solenoid is given by B = μ0 * N * I / l, where μ0 is the permeability of free space.

Now, let's look at the equation for power dissipated in a wire. We know that P = I^2 * R, where R is the resistance of the wire.

In order to show that the heat dissipated in the windings is independent of the size of the wire used, we need to show that the power dissipated is not dependent on the cross-sectional area of the wire.

To do this, let's first look at the resistance of a wire. We know that the resistance of a wire is given by R = ρ * l / A, where ρ is the resistivity of the wire, l is the length of the wire, and A is the cross-sectional area of the wire.

Now, let's substitute this into our equation for power dissipated:

P = I^2 * (ρ * l / A)

Next, let's substitute in our equation for the magnetic field inside the solenoid:

P = (μ0 * N * I / l)^2 * (ρ * l / A)

Now, let's simplify this equation:

P = μ0^2 * N^2 * I^2 * ρ / A

We can see that the cross-sectional area of the wire, A, is not present in this equation. This means that the power dissipated is not dependent on the size of the wire used.

I hope this helps! Let me know if you have any further questions.


 

1. What is the formula for calculating the heat dissipated in a solenoid wire?

The formula for calculating the heat dissipated in a solenoid wire is Q = I²Rt, where Q is the heat dissipated, I is the current flowing through the wire, R is the resistance of the wire, and t is the time.

2. How does the heat dissipated in a solenoid wire affect the temperature of the wire?

The heat dissipated in a solenoid wire will increase the temperature of the wire, as it is a form of energy that is converted into heat. The higher the heat dissipation, the higher the temperature of the wire will be.

3. What factors can affect the heat dissipated in a solenoid wire?

The factors that can affect the heat dissipated in a solenoid wire include the amount of current flowing through the wire, the resistance of the wire, the duration of the current flow, and the material of the wire.

4. How can the heat dissipated in a solenoid wire be minimized?

The heat dissipated in a solenoid wire can be minimized by using a wire with a lower resistance, reducing the current flow, or shortening the duration of the current flow. Additionally, using materials with high thermal conductivity can help dissipate the heat more efficiently.

5. What are the potential dangers of high heat dissipation in a solenoid wire?

High heat dissipation in a solenoid wire can lead to overheating and potentially damage the wire or other components in the circuit. It can also pose a fire hazard if the temperature becomes too high. Additionally, it can affect the accuracy and efficiency of the solenoid's operation.

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