5.7.2: Explorations

Exploration 1: A Long Wire with Uniform Current

The gray circle in the center represents a cross section of a wire carrying current coming out of the computer screen. The current is uniformly distributed throughout the wire (position is given in centimeters and magnetic field strength is given in millitesla). The black circle is an Amperian loop with a radius you can change with the slider. Restart.

Begin with the Amperian loop with a radius larger than the radius of the wire.

1. What is the radius of the Amperian loop?
2. What is the magnetic field at this radius?

You will use Ampere's law to find the total current in the wire:

\[\int\mathbf{B}\cdot d\mathbf{l}=\mu_{0}I,\quad\mu_{0}=4\pi\times 10^{-7}\text{ T}\cdot\text{m/A}\nonumber\]

where the integration is over a closed loop (closed path), \(d\mathbf{l}\) is an element of the path in the direction of the path, and \(I\) is the total current enclosed in the path. Pick a point on the Amperian loop and draw both the direction of the magnetic field at that point and the direction of \(d\mathbf{l}\) (tangent to the path).

3. The magnetic field and \(d\mathbf{l}\) should be parallel to each other. What is \(\mathbf{B}\cdot d\mathbf{l}\)?

Pick another point on the Amperian loop.

4. What is the magnitude of the magnetic field at that point? At any point on the loop?
5. This means you can write \(\int\mathbf{B}\cdot d\mathbf{l}=B\int dl\). Why?

\(\int dl\) is simply the length of the Amperian loop (in this case the circumference). Therefore, \(B =\mu_{0}I/2\pi r\) outside the wire.

6. Calculate the current carried by the wire from your measurement of the magnetic field.
7. Change the radius of the loop (but leave it still bigger than the wire) and predict the magnetic field on that loop. Measure the value to verify your answer.

Make the Amperian loop smaller so it fits inside the wire. This time, the current inside the loop is not equal to the total current, but instead it is equal to the total current times the fraction of the area inside the loop, \(Ir^{2}/a^{2}\), where \(a\) is the radius of the wire.

8. Why?
9. Use this fraction and Ampere's law again to predict the magnetic field inside the loop.
10. Measure the field to verify your answer.
11. Show that the general expression for the magnetic field inside the wire is \(\mu_{0}I r/2\pi a^{2}\).

Exploration authored by Anne J. Cox.

Exploration 2: A Plate of Current

Ampere's law states that \(\int\mathbf{B}\cdot d\mathbf{l}=\mu_{0}I\), where the integration is over a closed loop (closed path), \(d\mathbf{l}\) is an element of the path in the direction of the path, \(\mu_{0}\) is the permeability of free space (\(4\pi\times 10^{-7}\text{ T}\cdot\text{m/A}\)), and \(I\) is the total current enclosed in the path (position is given in millimeters and the magnetic field is given in millitesla \(10^{-3}\text{ T}\), so the integral is given in \(\text{mT}\cdot\text{mm} = 10^{-6}\text{ T}\cdot\text{m}\)). To use Ampere's law to calculate the magnetic field, Amperian loops need to mimic the symmetry of the field so that \(\mathbf{B}\cdot d\mathbf{l}\) is constant over the loop (or sections of the loop). Restart.

1. The blue dots represent wires carrying current into or out of the computer screen. In which direction does the current flow in the wires? Explain.

This animation shows the path integral (the value in table and on the bar graph) as you move the cursor (the circle with crosshair) around, as well as the position of the cursor as you move it. Move the cursor along the top portion of the loop.

2. Is the integral positive or negative? Why? (Hint: \(d\mathbf{l}\) points along the path in the direction you move the dot around the path).

Move the cursor to a corner and re-zero the integral (push the "set integral \(= 0\)" button). Now move the cursor along one of the vertical sides of the loop.

3. How does the size of this integral compare with the integral along the top part of the loop? Why? (Hint: What is the direction of \(\mathbf{B}\) along the side and what is the direction of \(d\mathbf{l}\)? So what then is \(\mathbf{B}\cdot d\mathbf{l}\)?)
4. Do the complete path integral (take the cursor completely around the loop). What is its value?
5. From this value, if each wire has the same current, what is the current in one of the wires?
6. From the path integral, what is the magnetic field above the series of wires? (Hint: If we neglect edge effects, \(\int\mathbf{B}\cdot d\mathbf{l}=BL\) on the top and bottom of the loop, and \(\int\mathbf{B}\cdot d\mathbf{l}=0\) for the sides.)
7. Compare your calculated value (from the path integral) with the value you measure by click-dragging around the animation. Comment on any differences.
8. Show that the general expression for the magnetic field above or below the series of wires is \(B = (\mu_{0}/2)\text{(current/length)}\) where the current/length is the current per length across the cross section of the plate (along the \(x\) axis in this animation).
9. Verify this expression for this animation.

Exploration authored by Anne J. Cox.
Script authored by Mario Belloni and Wolfgang Christian and modified by Anne J. Cox.

Exploration 3: Wire Configurations for a Net Force of Zero

The purple wire and the green wires have fixed currents and fixed positions. You can change the current in the gray wire by using the slider, and you can also drag the gray wire around to new positions. The animation shows the magnetic field vectors as well as the force on the wires. You can also add field lines by double clicking. Restart.

1. What are the directions of the currents in the green and purple wires?
2. Which one carries a larger magnitude of current? Explain.

Keep the current in the gray wire at zero (you can use the "\(I = 0\)" button to set the current to zero). Move the gray wire to a spot where the magnetic field is zero. Now increase the current in the gray wire.

3. Why is the force on the gray wire zero?
4. Why isn't the force on the other wires zero?
5. Is the force on the other wires different from the force on those wires before the current was turned on in the gray wire? Explain.

With current in the gray wire, move it to some point where the force is nonzero. The force on the wire is due to the current in the gray wire and the magnetic field it sits in (due to the other wires), \(\mathbf{F}=q\mathbf{v}\times\mathbf{B}=I\mathbf{L}\times\mathbf{B}\), where \(\mathbf{L}\) is the length of the wire and points in the direction of the current in the wire. To determine the direction of the force, then, you use the right-hand rule. Turn the current off in the gray wire.

6. What is the direction of the net magnetic field (make a sketch)?
7. Positive current comes straight out of the computer screen (negative current is into the screen). Therefore, in what direction is \(I\mathbf{L}\times\mathbf{B}\) for negative current (indicate this on your sketch)? Try it and verify your answer.
8. With a negative current, where does the gray wire need to be located so that the force on the purple wire is zero? So that the force on the green wire is zero? Explain.
9. If you change the current, how does your answer to (h) change? Explain.

Try a different configuration.

10. Where will the force be zero on the gray wire when it has a current flowing in it?
11. If the gray current has a current of about \(-1\text{ A}\), where do you have to put it in order to get the force on the green wire to be zero? Where do you have to put it in order to get the force on the purple wire to be zero? Where do you have to put it in order to get the force on the yellow wire to be zero? Explain.

Exploration authored by Anne J. Cox.

Physlets were developed at Davidson College and converted from Java to JavaScript using the SwingJS system developed at St. Olaf College.