What is the Magnetic Field of a Toroidal Solenoid?

In summary, the magnetic field at different distances from the center of a toroidal solenoid can be calculated using Ampere's Law, where the magnetic field is equal to the product of the permeability of free space, the number of turns in the solenoid, and the current enclosed within the loop divided by the circumference of the loop. For distances less than the inner radius, there is no current enclosed and thus no magnetic field. For distances between the inner and outer radius, there is a non-zero current enclosed and thus a non-zero magnetic field. For distances greater than the outer radius, the current enclosed is canceled out and thus the magnetic field is zero. The direction of the current and the magnetic field are also important factors in
  • #1
frostchaos123
17
0

Homework Statement



A toroidal solenoid has inner radius r1=15.0 cm and outer radius r2=18.0 cm. the solenoid has 250 turns and carries a current of 8.50 A. what is the magnitude of the magnetic field at the following distances from the center of the torus (a) 12.0 cm; (b) 16.0 cm; (c) 20.0 cm?


Homework Equations



Using Ampere's Law, B(2*pi*r) = [tex]\mu[/tex]NI

The Attempt at a Solution



The answer given is that if r < r1, the B is 0 as the I enclosed is 0.

However what i do not understand is that why is when r > r2, the I enclosed is also 0?
If r > r2, shouldn't the circle formed encompass the entire solenoid and thus all of the current?

Thanks.
 
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  • #2
No. its because again the total current THROUGH the area is 0. current going inwards through one turn is canceled by the current coming outwards in the next turn. try to imagine, did you get it?
 
  • #3
supratim1 said:
No. its because again the total current THROUGH the area is 0. current going inwards through one turn is canceled by the current coming outwards in the next turn. try to imagine, did you get it?

Not quite sure, but just to double confirm, is my representation of the amperian loop in red correct for r > r2? So do you mean that the current going into the page at the right hand side of the solenoid is canceled by the current coming out of the page on the left hand side?

[URL]http://img130.imageshack.us/i/solenoid.gif/[/URL]
 
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  • #4
no. every turn's upper part cancels the current of the lower part. got it?
 
  • #5
ok, i got it, apparently i got the magnetic field and the current flow mixed up.

So to summarise if the loop is bigger than r2, there will be a portion of the current flowing into the surface, and another equal portion coming out of the surface enclosed by the loop

And if the loop is between r1 and r2, then according to the diagram there will be just 1 portion of the current going into the surface and none out.
 

1. What is a solenoid?

A solenoid is a long, cylindrical coil of wire that is often used to create a magnetic field.

2. How is the magnetic field of a solenoid created?

The magnetic field of a solenoid is created by the flow of electric current through the coil of wire. This current creates a magnetic field that runs along the axis of the solenoid.

3. What is the direction of the magnetic field inside a solenoid?

The magnetic field inside a solenoid runs in a parallel direction to the axis of the solenoid, from the positive end to the negative end.

4. How is the strength of the magnetic field in a solenoid determined?

The strength of the magnetic field in a solenoid is determined by the number of turns in the coil of wire, the magnitude of the electric current flowing through the wire, and the material of the core (if present).

5. What are some practical applications of the magnetic field of a solenoid?

The magnetic field of a solenoid is used in a variety of devices, such as electric motors, generators, and speakers. It is also used in medical imaging machines, such as MRI scanners, to create detailed images of the body.

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