μ 0 I enclosed = B ds

Transcription

1 Ampere s law To determine the magnetic field created by a current, an equation much easier to use than Biot-Savart is known as Ampere s law. As before, μ 0 is the permeability of free space, 4π x 10-7 T m A I is electric current B is the magnetic field vector ds is a small step in space The integral symbol, summa, with a loop indicates a path integral, meaning the sum of all of the little steps must be a closed loop The right-hand side is a dot product, but all of the problems we do will have angles of either 0º or 180º between B and ds, so we only need to consider if the two are parallel or anti-parallel if we follow the convention that the ds steps are taken counter-clockwise. Suppose we have a wire which carries a current into the page, along the negative z-axis. We can use the cross product of Biot-Savart to determine the direction of the field at each of these four surrounding points. With the index finger into the page with the current flow, rotate the middle finger to point a and you will find your thumb points down. Rotate the middle finger to d and the thumb points left, to c and the thumb points up, to b and the thumb points right.

2 In general, the current creates a field that circulates around the wire clockwise. A simple shortcut for this is to point your right hand thumb with the current flow and see that your fingers naturally curl clockwise (the negative z-axis is tied to negative rotation in the xy-plane). This circulating magnetic field is true at any distance from the wire, though the strength is weaker with distance, so a fuller diagram would look like this:

3 Now let s apply Ampere s law by drawing a loop around the wire. The I enclosed is just the lone current, I, but along the negative z-axis, so I. If we look at the little step of ds on the right side, it is antiparallel to the magnetic field, so we have cos 180º from the dot-product, or -1. So now we have μ 0 (-I) = B ds ( 1) By radial symmetry, the magnetic field strength should be the same everywhere around the loop, so we can pull B out of the integral. We can pull out the -1 as well, leaving only ds. Conceptually, this is just the circumference of the loop, 2πR. This gives us μ 0 (-I) = B (-1) (2πR) So the magnitude of the magnetic field from a long, straight wire is: B = μ 0 I 2π R where R is the distance from the wire Answer Webassign Question 1 Answer Webassign Question 2

4 What if we had, instead of a thin wire, a current, I, of uniform density traveling through a cylinder of radius, a? The magnetic field for R > a is going to be what it was for the last example. The process of Ampere s law is exactly the same. What is the magnetic field for R < a? We ll simply draw an Amperian loop of radius R inside the cylinder. Now the enclosed current isn t the total current, I. It is only some fraction of that current, the fraction being the area within the green Amperian loop over the area of the entire cylinder. So I enclosed = I π R2 = I R2 π a 2 a 2 but, again, negative along the negative z-axis. The right-hand side of Ampere s law follows the same process as the previous example and will be B (2πR) (-1)

5 So μ 0 (- I R2 a2) = B (2πR) (-1) The magnitude of the magnetic field inside the cylinder is then B = μ 0 I 2π R a 2 This equation makes sense when R drops to zero and agrees with the previous equation when R = a. Answer Webassign Question 3 There are two common structures of coiled wire that can be used to create very strong magnetic fields, a solenoid and a torus. A solenoid is just a cylindrical coil of wire, something like an insulated Slinky. Let s say that looking at the solenoid from point P, you see a current I traveling clockwise in the solenoid. In this diagram, that means current is flowing out of the page on the top side of the solenoid and then flowing back into the page on the bottom side of the solenoid. Knowing that, let s draw a cut-away view.

6 Now let s take a look at one ring of the solenoid and what direction of magnetic field it would produce at point H. For the bottom of the ring, point your index finger into the screen with the current and rotate your middle finger towards H. You will see your thumb points down and right with the magnetic field. For the top of the ring, point you index finger out of the screen with the current and rotate your middle finger towards H. You will see your thumb points up and right with the magnetic field. The vertical components add to zero, leaving a field only to the left. All the other rings will have a similar effect at H, creating a very strong rightward field there. Likewise for any other point inside the solenoid. If you put H outside the solenoid, you can see the two magnetic fields almost add to zero. They get closer and closer to adding to zero as the coil radius R becomes smaller and smaller relative to the solenoid length, L. For an idealized solenoid, this external magnetic field is taken to be zero. So now let s apply Ampere s law to the red loop shown below.

7 We ve just stated that the magnetic field is essentially zero outside the solenoid, so B ds is going to be zero from a to b and also the portions of b to c and d to a outside the solenoid. But even for the portions of b to c and d to a which are inside the solenoid, the little steps of ds are perpendicular to the B field (which is we ve shown is pointing right), so B ds is going to be zero. So the only relevant path is from c to d. The enclosed current is the current in the solenoid, I, times the number of loops we ve included in the red box, which we will designate as N. B ds from c to d is just B x cos0º = B x So μ 0 I N = B x B = μ 0 I N x where N x is coils per length, often written as n For an ideal solenoid, B = μ 0 I n where n is coils per length Answer Webassign Question 4 The second common geometry of a coiled wire is a torus or toroid. This is basically a solenoid which has had the two ends connected to form the shape of a bagel. If we imagine that the current flows into the screen along the outer radius of the toroid and then out of the screen on the inner radius of the toroid, the cut-away view would look like this:

8 We ll draw an Amperian loop inside the toroid with a radius R and assume the toroid has N number of turns and carries a current, I. I enclosed is simply I N The loop in enclosing a current out of the page and if you point your right hand thumb with that current, your fingers curl counter-clockwise with the magnetic field. So the dot product will involve cos0º. And the sum of the ds steps is just the circumference of the Amperian loop. B ds = B (2π R) (cos0º) = B 2πR μ 0 I N = B 2πR So for a toroid, B = μ 0I N 2π R where N is the number of turns in the toroid Answer Webassign Question 5