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Question Video: Calculating Torque on a Current-Carrying Loop of Wire in a Uniform Magnetic Field Physics

The diagram shows a rectangular loop of current-carrying wire between the poles of a magnet that produces a field with a magnitude of 250 mT. The longer sides of the loop are initially parallel to the magnetic field, and the shorter sides of the loop are initially perpendicular to the magnetic field. The loop has a magnetic dipole moment of 500 ΞΌN β‹… m/T. The loop is then rotated through 90Β° so that all of its sides are perpendicular to the magnetic field. How much does the torque on the loop change by due to its rotation? Answer to the nearest micronewton-meter.

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Video Transcript

The diagram shows a rectangular loop of current-carrying wire between the poles of a magnet that produces a field with a magnitude of 250 millitesla. The longer sides of the loop are initially parallel to the magnetic field, and the shorter sides of the loop are initially perpendicular to the magnetic field. The loop has a magnetic dipole moment of 500 micronewton-meters per tesla. The loop is then rotated through 90 degrees so that all of its sides are perpendicular to the magnetic field. How much does the torque on the loop change by due to its rotation? Answer to the nearest micronewton-meter.

The rectangular current-carrying loop in this scenario starts out in this orientation between the poles of our magnet. In this orientation, the wire experiences a torque given by this equation. The wire, though, doesn’t stay still but instead rotates 90 degrees until it reaches this vertical orientation. Arranged that way, the wire also experiences some amount of torque. We want to find the change in torque experienced by the wire due to this 90-degree rotation. If we call the torque acting on the wire in its initial position 𝜏 one and the torque on the wire after it’s been rotated 90 degrees 𝜏 two, what we’re after is the magnitude of the difference between these values.

Returning to our general equation for the torque on a current-carrying wire in a magnetic field, we note that the values on the right-hand side are magnetic field 𝐡, current 𝐼, cross-sectional area of the loop 𝐴, number of turns of the loop 𝑁. And all this is multiplied by the sine of an angle called πœƒ. Here’s how πœƒ is defined. If we have our current-carrying loop, in this case a rectangle, and we draw a vector that is normal or perpendicular to the area of that loop, then the angle between that vector and the external magnetic field is πœƒ.

Knowing this, we can now express this magnitude in terms of this equation for torque, writing an expression for 𝜏 one. This will be equal to the magnetic field strength times the current 𝐼 times the area 𝐴 of the loop multiplied by the number of turns in the loop. But then, working on our diagram, we see that in this case 𝑁 is equal simply to one. All this is multiplied by the sine of an angle that we’ll leave for now as πœƒ one. So 𝜏 one equals 𝐡 times 𝐼 times 𝐴 times the sin of πœƒ one. And note that we haven’t put subscripts on these variables here because they’re constant regardless of the orientation of our coil. This means that when we write an expression for 𝜏 two, we use the same values for 𝐡 and 𝐼 and 𝐴. The only thing that’s different is that now we have an angle πœƒ two.

So then, the magnitude of 𝜏 one minus 𝜏 two equals the magnitude of 𝐡𝐼𝐴 sin πœƒ one minus 𝐡𝐼𝐴 sin πœƒ two. We can factor out at 𝐡 times 𝐼 times 𝐴 from the interior of this expression. At this point, let’s figure out what these angles πœƒ one and πœƒ two are. πœƒ one refers to an angle when our coil is in this initial horizontal orientation. It’s equal to the angle between a vector that is perpendicular to the plane of this loop and the magnetic field created by this permanent magnet, which goes from the North Pole to the South Pole. We can see that these two vectors cross at 90 degrees. Therefore, that’s the value of πœƒ one.

When we go to solve for πœƒ sub two, now we look at the case where our coil is oriented vertically. The magnetic field the coil exists in points in the same direction as before. But now, a normal vector, or a perpendicular vector, to the plane of our coil points antiparallel to 𝐡. πœƒ two then is equal to 180 degrees. If we plug our values for πœƒ one and πœƒ two into our expression, we’re now evaluating a term where we have the sin of 90 degrees minus the sin of 180 degrees. The sin of 90 degrees, though, equals one, and the sin of 180 degrees equals zero.

Note that if instead of 180 degrees we had zero degrees here, that is, if the area vector of our loop pointed to the left instead of to the right, we would get the same answer. The sin of zero degrees equals the sin of 180 degrees, which is zero. The change in 𝜏 then equals the magnitude of 𝐡 times 𝐼 times 𝐴 times the quantity one minus zero or simply 𝐡 times 𝐼 times 𝐴, where all these numbers are positive values.

To move forward on our solution, we’ll need to recall the information given to us in the problem statement. We were told that the magnetic field magnitude 𝐡 is 250 millitesla and that something called the magnetic dipole moment of the loop, represented by the Greek letter πœ‡, is 500 micronewton-meters per tesla. In general, the magnetic dipole moment of a current-carrying loop is equal to the current that is carried by the loop multiplied by its cross-sectional area. Therefore, our expression 𝐡 times 𝐼 times 𝐴 can equivalently read 𝐡 times πœ‡, the magnetic dipole moment.

We substitute in the values for 𝐡 and πœ‡. And before we multiply them together, let’s look at the value for 𝐡 and convert it from millitesla into tesla. We do this so that the units of 𝐡 can cancel out with the units of inverse tesla in our value of πœ‡. Since there are 1000 millitesla in one tesla, that means to convert millitesla to tesla, we move the decimal place three spots to the left.

Now we have a magnetic field strength in units of tesla. And those units cancel out with the units of inverse tesla in our value for πœ‡. Note that this leaves us with final units of micronewton-meters. We want to give our final answer to the nearest micronewton-meter. Calculating this product, we get a result of 125 micronewton-meters. This is how much the torque on our current-carrying loop changes due to its rotation.

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