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Question Video: Determining Magnetic Field Magnitude for a Current-Carrying Loop of Coil Physics

The diagram shows a rectangular conducting coil with 3 turns that is in a magnetic field. There is a current of 8.5 A in the coil. The sides of the loop parallel to line 𝑑₁ are parallel to the magnetic field, and the sides of the loop parallel to line 𝑑₂ are perpendicular to the magnetic field. The length of 𝑑₁ = 0.035 m and the length of 𝑑₂ = 0.025 m. The torque on the loop is 15 mNβ‹…m. Find the magnitude of the magnetic field to the nearest millitesla.

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Video Transcript

The diagram shows a rectangular conducting coil with three turns that is in a magnetic field. There is a current of 8.5 amperes in the coil. The sides of the loop parallel to line 𝑑 one are parallel to the magnetic field, and the sides of the loop parallel to line 𝑑 two are perpendicular to the magnetic field. The length of 𝑑 one equals 0.035 meters, and the length of 𝑑 two equals 0.025 meters. The torque on the loop is 15 millinewton-meters. Find the magnitude of the magnetic field to the nearest millitesla.

In our diagram, we see our rectangular conducting coil with three turns or loops in it. This coil is positioned between the poles of a permanent magnet. For this reason, the coil is exposed to a constant magnetic field that we’ll call 𝐡. In this example, we want to solve for the magnitude of that field to the nearest millitesla. To begin doing this, let’s clear some space on screen and recall that the general equation for the torque acting on a current-carrying loop in a magnetic field is given by this expression. Here, the magnitude of that magnetic field is 𝐡, the current in the conducting loop is 𝐼, the cross-sectional area of the loop is 𝐴, and 𝑁 represents the number of turns in the conducting coil. Along with all this, we see that there’s an angle of πœƒ involved in this expression.

To understand πœƒ, imagine that we have a rotating coil and a magnetic field, and we draw a vector that’s perpendicular to the plane of the rotating coil. The angle between that vector and the external magnetic field, this angle here, that is πœƒ. We can see then that πœƒ will change as our rotating coil turns.

If we look at our given scenario, though, we see that in this orientation a vector that is perpendicular or normal to the plane of our conducting coil looks like this. We see that the angle between this vector and the external magnetic field is 90 degrees. As we look to apply this general equation then, in our case, we can write that the torque 𝜏 equals 𝐡 times 𝐼 times 𝐴 times 𝑁 times the sin of 90 degrees. The sin of 90 degrees equals one. So, we can effectively leave out this factor from our expression.

Let’s recall that it’s not the torque we want to solve for, but rather the magnetic field magnitude. To do that, we can divide both sides of the equation by 𝐼 times 𝐴 times 𝑁, canceling those factors on the right. The magnetic field strength 𝐡 equals 𝜏 divided by 𝐼 times 𝐴 times 𝑁. At this point, we can recall the information given to us in our problem statement. We were told for one thing that at this moment in time the torque acting on our conducting coil, we’ll call it 𝜏, is equal to 15 millinewton-meters. As we work with this term, we’ll just need to be careful to recall that this β€œm” indicates milli-, the prefix meaning one thousandth, while this β€œm” indicates the unit meters.

We were also told the current, we’ll call it 𝐼, that exists in our conducting coil. It’s 8.5 amperes. Regarding the cross-sectional area 𝐴 of our coil, we weren’t given that value, but we were told the dimensions, 𝑑 one and 𝑑 two, of this rectangular coil. We can multiply those together, and that will equal the area 𝐴. Lastly, our coil has more than one loop to it. We’ll call that number capital 𝑁. And we were told that it’s equal to three. We can even see these three turns in the coil in our diagram. Knowing all this, we’re now ready to substitute into our equation and begin solving for 𝐡.

Note that as we do this, for the variable 𝐴, we put in 𝑑 one multiplied by 𝑑 two. The product of those two values will give us the cross-sectional area 𝐴. Thanks to the prefix milli- in the numerator of our fraction, if we calculate this expression as it’s written, we’ll get an answer in units of millitesla; that’s just what we want. And when we calculate this answer, we get a result of 672.268 and so on millitesla. And now we recall that we want to round our result to the nearest millitesla. Since the first decimal place digit is less than five, our final answer is 672 millitesla. This is the magnitude of the magnetic field surrounding our current-carrying conductor.

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