Question Video: Calculating Torque and Magnetic Dipole Moment for 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. The sides of the loop parallel to line ๐‘‘โ‚ are parallel to the magnetic field, and the sides of the loop parallel to line ๐‘‘โ‚‚ are perpendicular to the magnetic field. The current in the loop is 350 mA, and the magnetic field strength is 0.12 T. The length of ๐‘‘โ‚ = 0.025 m and the length of ๐‘‘โ‚‚ = 0.015 m. Find the torque acting on the loop to the nearest micronewton-meter. Find the magnetic dipole moment of the loop to the nearest micronewton-meter per tesla.

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Video Transcript

The diagram shows a rectangular loop of current-carrying wire between the poles of a magnet. The sides of the loop parallel to line ๐‘‘ one are parallel to the magnetic field, and the sides of the loop parallel to line ๐‘‘ two are perpendicular to the magnetic field. The current in the loop is 350 milliamperes, and the magnetic field strength is 0.12 tesla. The length of ๐‘‘ one equals 0.025 meters, and the length of ๐‘‘ two equals 0.015 meters. Find the torque acting on the loop to the nearest micronewton-meter.

Looking at our diagram, here is our current-carrying loop in a uniform magnetic field between the poles of this magnet. Because thereโ€™s current in all four sides of our loop, there is potentially a magnetic force on each of these four sides. In our problem statement though, we were told that the two sides of the rectangle parallel to this distance ๐‘‘ one, so that would be this side here and this side here, are both exactly parallel to the external magnetic field. Therefore, the current in these two sides of the rectangle is either parallel or antiparallel with the external field. And that means these sides of the rectangle experience no magnetic force. Itโ€™s only the two sides perpendicular to that field, this side and this side of the rectangle, that experience a nonzero magnetic force.

We can determine the direction of that force on each of these side lengths according to whatโ€™s called a right-hand rule. Given a wire carrying current weโ€™ll call ๐ผ, in a uniform magnetic field weโ€™ll call ๐ต, then using our right hand, if we point the index finger of that hand in the direction of ๐ผ and then the middle finger of that hand in the direction of ๐ต, that will make the thumb of our right hand point in the direction of the resultant magnetic force. That is, the force on this segment of current-carrying wire will be out of the screen towards us. This right-hand rule will let us figure out the direction of the force acting on either side of our rectangle thatโ€™s perpendicular to the external magnetic field.

On the side of the rectangle thatโ€™s to the left as we see it, this side here, pointing the index finger of a right hand in the direction of current here and the middle finger of that hand in the direction of the external magnetic field, left to right, the right-hand rule tells us that the resulting magnetic force direction is down. Considering next the opposite side of our rectangle, in this side, current points in this direction. Like before, the external magnetic field points left to right so that now by our right-hand rule the force on this segment of the rectangle is upward.

The combined effect of these two forces is to create a torque. There will be a tendency of our coil of wire to rotate about an axis through the loopโ€™s center. This is the torque our question is asking about. Thereโ€™s a mathematical relationship for the torque on a current-carrying wire in a magnetic field that we can recall. That torque ๐œ is equal to the magnetic field strength ๐ต times the current in the wire ๐ผ multiplied by the cross-sectional area of the wire ๐ด times the number of turns in the wire multiplied by the sine of an angle weโ€™ve called ๐œƒ.

Before weโ€™re able to evaluate this expression for our scenario though, letโ€™s recall the information given to us in our problem statement. Our current-carrying loop, we remember, has dimensions ๐‘‘ one multiplied by ๐‘‘ two. ๐‘‘ one is given as 0.025 meters, while ๐‘‘ two is 0.015 meters. If we were to write a specific equation then for the torque on our given current-carrying loop, we can replace the cross-sectional area ๐ด of the loop with ๐‘‘ one multiplied by ๐‘‘ two. Another bit of information given to us in our problem statement is that the current carried by our loop is 350 milliamperes, what weโ€™ll call ๐ผ. Along with this, weโ€™re told that the magnitude of the magnetic field is 0.12 tesla. Weโ€™ll call this ๐ต.

In this equation for torque, the capital ๐‘, as we mentioned, refers to the number of turns in our loop. In this scenario, ๐‘ is equal to one, and so we can drop it out of this equation. When it comes to the sin of this angle ๐œƒ, we can note that ๐œƒ is the angle between the external magnetic field and the area vector of a rectangular loop. Based on the direction of current in the loop, that area vector would point directly downward. Notice this means that the angle between this area vector and the external field is 90 degrees. When we evaluate the sin of 90 degrees, we get one. And we can again drop out this factor. It wonโ€™t have an effect on the final torque acting on the loop.

For our next step, we can substitute these known values into our equation for torque. In place of ๐ต, we put 0.12 tesla. In place of the current ๐ผ, instead of 350 milliamperes, we convert this to a number in amperes, 0.350. Note that weโ€™ve shifted the decimal place three spots to the left to move from milliamperes into amperes. Then, ๐‘‘ one is 0.025 meters and ๐‘‘ two is 0.015 meters. In scientific notation, the exact answer we get is 1.575 times 10 to the negative fifth newton-meters. In terms of units, note that one newton is a tesla ampere meter.

So then we have an exact answer in newton-meters. But our question asks us to give an answer to the nearest micronewton-meter. A micronewton is one millionth of one newton, meaning that if we wrote our answer as simply a decimal number, we could convert it into micronewton-meters by moving the decimal place one, two, three, four, five, six places to the right. We have then 15.75 micronewton-meters. Weโ€™re almost there. We just need to round it to the nearest whole number. The closest whole number to 15.75 is 16, which means that, to the nearest micronewton-meter, the torque acting on this coil is 16 micronewton-meters.

Now, letโ€™s look at part two of our question.

Find the magnetic dipole moment of the loop to the nearest micronewton-meter per tesla.

If we have a loop of current-carrying wire, where the current has a magnitude ๐ผ and the area of that loop is called ๐ด, then the magnitude of the magnetic dipole moment of the loop ๐œ‡ is equal to ๐ผ times ๐ด. To solve for ๐œ‡ in our case, weโ€™ll use the fact that current in our rectangular loop is 350 milliamperes and that ๐‘‘ one and ๐‘‘ two, the perpendicular dimensions of our rectangular loop, are given.

Substituting these values into our equation, notice that the units we have are milliamperes and meters times meters, or meters squared. We want our final answer though in units of micronewton-meters per tesla. Recalling that one newton equals a tesla ampere meter, we can rearrange this to see that an ampere equals a newton per tesla meter. All this means we can replace the ampere here in our equation by a newton per tesla meter. And note that this m in our equation is not actually the unit of meter but is the prefix milli-. Really, what we have is millinewtons per tesla meters.

And then recall that weโ€™re multiplying this unit by the units of meters squared. When we multiply all these numbers together, we get a numerical value of 0.13125, where the units are millinewton-meters per tesla. Those arenโ€™t exactly the units of our final answer though. We want them to be in micronewton-meters per tesla. To convert between these two, we can recall that one millinewton equals 1000 micronewtons. This means that to change our units to micronewton-meters per tesla from millinewton-meters per tesla, weโ€™ll need to multiply this number here by 1000. Doing that gives us a result of 131.25 micronewton-meters per tesla.

Lastly, weโ€™ll report our final, final answer to the nearest micronewton-meter per tesla. Rounding our answer to the nearest whole number, we report that the magnetic dipole moment of the loop to the nearest micronewton-meter per tesla is 131.

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