Magnetic field with solenoid, have work;

In summary: B=μnI , I=B/μnn=N/L=772/.772=1000 turns per lengthI=B/μn = 8e-3/(μ*1000) = 6.3662AP=I2R=212.77W
  • #1
Th3Proj3ct
19
0

Homework Statement



A solenoid 10.0 cm in diameter and 77.2 cm long is made from copper wire of diameter 0.100 cm, with very thin insulation. The wire is wound onto a cardboard tube in a single layer, with adjacent turns touching each other. What power must be delivered to the solenoid if it is to produce a field of 8.00 mT at its center?

Homework Equations


Alright, I know
B=μnI , I=B/μn
n=N/L
R=ρl/A
l=2πr(N)
A=π*radius(wire)^2
and P=I2R

The Attempt at a Solution



I've gotten all these equations, but just can't put them together >_<
n=N/L=772/.772=1000 turns per length
I=B/μn = 8e-3/(μ*1000) = 6.3662A

now just for R,
R=ρl/A
l=2πrN = 2(3.1415)(.05)(772) = 242.53
A=π*radius(wire)^2= (3.1415)(.0005)^2 = 7.854e-7
R=(1.7e-8*0.314159)/(7.854e-7)
= 5.25

So, P = RI^2 = 5.25*6.3662^2 = 212.77W !
 
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  • #2
Th3Proj3ct said:

The Attempt at a Solution



I've gotten all these equations, but just can't put them together >_<
n=N/L=.772m/.0001m=772turns per length
I=B/μn = 8e-3/(μ*772) = 8.2464A

772 is not your turns per length, it is your total number of turns for that solenoid.
 
  • #3
Snazzy said:
772 is not your turns per length, it is your total number of turns for that solenoid.

Was that how I was supposed to figure that out? The number of turns was one of the biggest issues I had in this problem.

n is supposed to be the turns per length though, right? And to do that I should take the total number of turns divided by the overall length, which would be 772/(2πr) or 772/0.3141592?
 
  • #4
The number of turns you have for a solenoid is a unitless number. You did the right thing by finding the total number of turns by dividing the length of your solenoid by the diameter of the wire. Good. Now you have 772 turns. This is your capital N variable. If you want to find small n, i.e., the number of turns per length, you'd have to divide the number of turns you have by the length of your solenoid. The length of your solenoid is not [tex]2\pi r[/tex].
 
  • #5
small n is what's needed in the function for the magnetic force(B) though, and if I do that, I just get 772/.772 (if 77.2cm is the length), so the little n is 1000?

btw is the other stuff looking right?

Edit: And thanks for all the help, this problem has been driving me crazy.
 
  • #6
It doesn't matter what formula you use. You could use:

[tex]B=\frac{ \mu_0 N I}{L}[/tex]

or:

[tex]B= \mu_0 n I[/tex]

where:

[tex]n=\frac{N}{L}[/tex]

Everything else seems to be fine. I believe, however, that the length of the copper wire would be the circumference of one wire multiplied by the total number of wires.
 
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  • #7
Snazzy said:
It doesn't matter what formula you use. You could use:

[tex]B=\frac{ \mu_0 N I}{L}[/tex]

or:

[tex]B= \mu_0 n I[/tex]

where:

[tex]n=\frac{N}{L}[/tex]

Everything else seems to be fine. I believe, however, that the length of the copper wire would be the circumference of one wire multiplied by the total number of wires.

As I said before, thanks a ton; it all really helped. I was able to finally get the right answer and yes you were right [i edited the original post with the correct work, giving the correct answer. Very much appreciated!
 

FAQ: Magnetic field with solenoid, have work;

What is a solenoid?

A solenoid is a coil of wire that is tightly wound in a helical shape. When an electric current flows through the wire, it creates a magnetic field.

How does a solenoid create a magnetic field?

The tightly wound coil of wire in a solenoid creates a concentrated and uniform magnetic field when an electric current flows through it. This is because the closely spaced loops of wire all contribute to the overall magnetic field in the same direction.

What is the purpose of a solenoid in a magnetic field?

Solenoids are used to create and control magnetic fields. They are commonly used in devices such as electromagnets, relays, and electric motors.

How does a magnetic field with solenoid work?

A magnetic field with solenoid works by passing an electric current through the coil of wire, which creates a magnetic field around the solenoid. This magnetic field can then be used to exert a force on other magnetic objects or to induce an electric current in nearby conductors.

What factors affect the strength of a magnetic field with solenoid?

The strength of a magnetic field with solenoid is affected by the number of turns in the coil, the amount of current flowing through the wire, and the material of the core inside the coil. The closer the turns are to each other and the higher the current, the stronger the magnetic field will be.

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