Video Transcript
The diagram shows a square section of wire that has been positioned in a uniform magnetic field such that two of its sides are perpendicular to the direction of the field and the other two sides are parallel to the field. The magnetic field has a strength of 0.2 teslas and the current through the wire is five amperes. Each side of the square is 0.1 meters long. What is the torque exerted on the wire by the magnetic field?
This square section of wire we see carries a counterclockwise current, and it exists in a magnetic field that points to the left. We know that the current in this square section of wire consists of individual charges. An electric charge when it’s moving through a magnetic field experiences a magnetic force. There is an exception to this rule, though. If an electric charge happens to be moving in the same direction as a magnetic field or exactly opposite the field, then in those two cases only it experiences no force. That fact is relevant to us because at the top of our square, we do have charge that moves parallel to the magnetic field and then at the bottom of our square we have charge moving opposite, or antiparallel, to that field.
What we’re saying is that the electric charges as they move through these two sides of our square do not experience a magnetic force. However, when electric charge moves through either the left or the right side of our square, it does experience a force. Clearing some space on screen, we note that the direction of the magnetic force acting on a charged particle moving through magnetic field is given by something called the right-hand rule. The way this rule works if we have a charge 𝑞 moving with a velocity 𝑣 within a uniform magnetic field we’ll call 𝐵, then if we point the fingers on our right hand in the direction of 𝑞 times 𝑣 and then curl those fingers so they point in the direction of the magnetic field 𝐵, arranged this way, the thumb of our right hand will point in the direction of the force on the charge 𝑞.
Let’s now use the right-hand rule on both the left and the right sides of our square of wire. Starting with the left side, we know that current travels downward on this side of the square, and conventionally current represents the flow of positive charge. For this side of the square then, 𝑞 times 𝑣 points downward. The magnetic field that our square of wire is in points to the left. So to follow the right-hand rule, we take our right hand and point our fingers downward. And from there we curl them so that they point to the left. Arranging our right hand so that this movement is possible means that the thumb of our right hand must point into the screen. Therefore, on the left side of our square, the force acting on the wire points into the screen.
If we then think about the force on the right side of the wire, here our current moves to the top of the screen. Therefore, 𝑞 times 𝑣 points to the top of the screen. And just like before, the magnetic field points to the left. Now, if we arrange our right hand so that our fingers can point upward and then can curl in the direction of the magnetic field, our hand is arranged so that our right thumb is pointing out of the screen towards us. Therefore, on the right side of the square, the magnetic force points out of the screen. We can now see how this force creates a torque on this square of wire. If we were to sketch in an axis through the center of the wire, then these two forces would tend to create a rotation about this axis like this. This is the torque that we want to calculate.
And to begin doing it, let’s again clear space at the top of our screen. And let’s recall that in general, the torque 𝜏 acting on an object is equal to the force acting on that object multiplied by the distance between the line of action of the force and the axis of rotation being considered. In the case of our square of current-carrying wire, we have this force here acting on the left side and this force acting on the right. Because these forces acting opposite directions, we can say that the total force, what shows up as 𝐹 in our equation for torque, is really equal in our case to two times 𝐹. We’re adding together the force on the left and the right side of our square.
Now let’s say that we call the length of the side of our square 𝑙. The distance between the line of action of either one of our two forces and the axis of rotation that runs through the center of the square is half of the length 𝑙, or 𝑙 over two. If we call the torque that we want to solve for 𝜏, then we now have an equation for that torque. However, we don’t yet know the magnitude of the force 𝐹 that acts on either side of our square. To help us with this, we can recall the general equation that the magnetic force 𝐹 on a current-carrying section of wire of length 𝑙 carrying a current of magnitude 𝐼 in a magnetic field 𝐵 is equal to 𝐵 times 𝐼 times 𝑙.
This equation holds true when the wire and the magnetic field are perpendicular to one another, as they are on both the left and the right sides of our square. Our equation for torque then becomes two times 𝐵 times 𝐼 times 𝑙 times 𝑙 over two. Notice that the factor of two cancels out with the factor of one-half, and we have two factors of 𝑙. So our equation becomes 𝐵 times 𝐼 times 𝑙 squared. Our magnetic field strength is given as 0.2 teslas, our current is five amperes, and the length of the side of the square is 0.1 meters. Calculating this result, we get an answer of exactly 0.01 newton-meters. This is the torque exerted on the wire by the magnetic field.
