Helmholtz coils are two identical circular coils having the sa...

Question

Helmholtz coils are two identical circular coils having the same radius 𝑅 and the same number of turns N, separated by a distance equal to the radius 𝑅 and carrying the same dc current I in the same direction. (See Fig. 28–61.) They are used in scientific instruments to generate nearly uniform magnetic fields. (They can be seen in the photo, Fig. 27–19.) (a) Determine the magnetic field B at points 𝓍 along the line joining their centers. Let 𝓍 = 0 at the center of one coil, and 𝓍 = 𝑅 at the center of the other. (b) Show that the field midway between the coils is particularly uniform by showing that dB/d𝓍 = 0 and d²B/d𝓍² = 0 at the midpoint between the coils. (c) If 𝑅 = 10.0 cm, N = 85 turns and I = 3.0 A, what is the field at the midpoint between the coils, 𝓍 = 𝑅/2 ?

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Accepted Answer

Hi, everyone. Let's take a look at this practice problem dealing with magnetic fields. This problem says in an experiment with Ham holds coils, each coil has a radius of 12.0 centimeters, 150 turns and they are separated by a distance equal to the radius. If the coils are carrying a current of 1.5 amps in the same direction, determine the magnetic field at the 0.6 0.0 centimeters from the center of one of the coils along the line joining their centers. Below the question, we're given a diagram of what was described in the problem. We're also given four possible choices as our answers. For choice A we have 1.7 milli tesla Ihat. Choice B. We have 2.8 milli tla Ihat choice C. We have 2.8 million tesla J hat. And for choice D, we have 4.1 milli tesla J hat. Now the current in each of the coils is going to generate a magnetic field and we actually know the magnetic field for a one of the single coils. So recall your magnetic field for coil about its axis through the center of that coil. And so that magnetic field is going to be B is equal to N multiplied by U mu not multiplied by I multiplied by R squared divided by the quantity of two multiplying the quantity of R squared plus X squared in quantity raised to the three halves. That fraction is multiplied by I hat where B here is our magnetic field. N is the number of turns in the coil. Muno is the probability of free space I is the current R is the radius of the coil and X is the distance from the center of the coil to the point where we're measuring the magnetic field. Now, here we chose the I hat direction. That's because if I look at the direction the current is flowing and I use my right hand rule, my magnetic field is going to be pointing to the right. And based upon the quar system that we're given in the diagram that's in the positive X direction, which would be the I hat direction. Now, this is the magnetic field only for one of the coils, but we actually have two coils. So we'll need to add their magnetic fields together to get our total magnetic field. So we'll have B total is equal to, for our first magnetic field, we're gonna just use our same formula which is gonna be N multiplied by Muno multiplied by I multiplied by R squared divided by the quantity of two multiplying the quantity of R squared plus X squared in quantity raised to the three halves and that's multiplied by I hat. Then we'll have to add to this the coil, the magnetic field due to the second coil. And here, the only difference is we're gonna replace X with R minus X. Since X is the distance from the center of the say left coil in our diagram and the coils are separated by distance R, then the distance between the point we're looking at and the second coil is just gonna be R minus X. So plugging in that substitution, we're gonna have in multiplied by Muno multiplied by I multiplied by R squared divided by the quantity of two, multiplying the quantity of R squared plus the quantity of R minus X in quantity squared. Then another in quantity raised to the three halves and then that fraction is again multiplied by I hat. So now all we need to do is add these two together. And before I plug in values, I notice that I could factor out a lot of the terms. So we're gonna do that, we'll have B total is equal to and I can factor out that N Muno ir squared divided by two. So I'll have N multiplied by Muno multiplied by I multiplied by R squared all divided by two. And that's gonna be multiplying the quantity of one divided by the quantity of R squared plus X squared in quantity raised to the three halves, then we'll have plus one divided by the quantity of R squared plus the quantity of R minus X in quantity squared. Another in quantity raised to three halves. And that whole thing is multiplied by, by hat. So now we can actually plug in values. So we'll have B total is equal to, for N, that is the number of turns. So that's gonna be 150. This can be multiplied by Muno which has value of four pi multiplied by 10 to the negative seven tesla meters per amp. That's gonna be multiplied by our current, which is 1.5 amps. And that could be multiplied by a radius R which is 12 centimeters. But I'll need to convert that into meters by dividing by 100. So it'll be 0.12 m and that radius is actually squared. Then that's gonna be divided by two. And then that's gonna be multiplying the quantity of one divided by quantity of R squared, which is the 0.12 m. That's gonna be squared plus X which we're looking at the midpoint which is six centimeters. So again, divided by 100 to convert centimeters into meters. That's gonna be 0.06 m. And that can be squared and that quantity is raised to the three halves, then we'll have plus one divided by the quantity of again R squared, which is the 0.12 m in quantity squared plus the quantity of R minus X. So let's be 12 centimeters minus six centimeters, which gives me six centimeters. So it could be 0.06 m and that is squared. Then that whole quantity is raised to the three halves and then everything is multiplied by I hat. So now we can plug our values to our calculator. I'm get B total is equal to 1.7 multiplied by 10 to the negative three teslas. And this is going in the I hat direction. So here I've just rounded to two significant figures. However, if I look at my answer choices, they're all in milli tesla. So I'll need to multiply this by the conversion factor of 1000 mi tesla divided by one tesla. And so recalculating, we'll get B total is equal to 1.7 milli tesla in the I hat direction. And so that is the answer that I'm looking for. And if I look at my answer choices, this corresponds to answer choice. A. So the trick to this problem was just to recall your formula for the magnetic field for a coil of wire where that magnetic field is along the central axis of that coil. Here, we actually had two coils. So we need to add their magnetic fields together to get our total magnetic field. And then we just had to plug in the values that we were given in the problem. So I hope that this has been useful and I'll see you in the next video.

Key Concepts

Magnetic Field of a Coil

Superposition Principle

Uniform Magnetic Field

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