# Question Video: Calculating the Applied Magnetic Field That Produces a Known Hall Voltage For a Known Current Flow

The Hall voltage measured by a Hall probe for a certain sample is 3.5 𝜇V when a current of 2.7 A and a magnetic field magnitude of 1.2 T is used to generate the Hall voltage. If a current of 2.1 A generates a Hall voltage measurement of 3.0 𝜇V for the same sample, what is the magnitude of the magnetic field?

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### Video Transcript

The Hall voltage measured by a Hall probe for a certain sample is 3.5 microvolts when a current of 2.7 amps and a magnetic field magnitude of 1.2 tesla is used to generate the Hall voltage. If a current of 2.1 amps generates a Hall voltage measurement of 3.0 microvolts for the same sample, what is the magnitude of the magnetic field?

In this scenario, we can imagine we have a sample of material that looks something like this. And this sample of material is in a magnetic field, we can call it capital 𝐵, which points perpendicular to the largest cross-sectional area of that sample. In addition, current, we can call it capital 𝐼, travels along the length of this sample. And under these conditions, as oppositely charged particles accumulate on the top and bottom of this sample, a Hall voltage across that distance is created. If we refer to the Hall voltage as 𝑉 sub 𝐻, then we can write down information for that voltage, the magnetic field and the corresponding current.

In the first example described in our statement, the Hall voltage is 3.5 microvolts. That corresponds to a current of 2.7 amps and a magnetic field of 1.2 tesla. Then, in a separate example, we have a current of 2.1 amps and a measured Hall voltage of 3.0 microvolts. And we wanna solve for the magnetic field 𝐵 in this second case. We’re looking, then, for some mathematical relationship that ties together these three variables, Hall voltage, magnetic field, and current. Here is a mathematical relationship that does that, that connects these three values.

We see the Hall voltage, 𝑉 sub 𝐻, is equal to the current through the sample multiplied by the magnetic field strength that’s in multiplied by 𝑑, where 𝑑 is the distance from the top of the sample to the bottom, divided by 𝑛, the number density of charge carriers, in this case electrons in the current, multiplied by the charge of each of those charge carriers times 𝐴, which is the cross-sectional area of our sample. And here on our diagram, we show what particular cross section that is. We don’t know all the values for the terms in this expression. But given the information we have in these two examples, we have enough to solve for the unknown magnetic field 𝐵.

To solve for the unknown magnetic field, we can use our equation for Hall voltage in each of our two example cases. Then, if we take the ratio of one instance to another, we’ll have what we need to solve for the unknown magnetic field. Let’s start doing this by applying this equation for Hall voltage to the first set of information we have, the first row in this table we’ve made. And to do that, let’s clear some space on screen.

Now, what we’re going to do is use this equation for Hall voltage and fill in all the information we can, based on what we know. We can say that 3.5 microvolts is equal to 2.7 amps times 1.2 tesla times 𝑑, which we don’t know, divided by 𝑛 times 𝑞 times 𝐴, none of which we know for certain. Now, really, we could reasonably assume that 𝑞 with the charge on the charge carriers is the magnitude of the charge of an electron. But, as we’ll see, we don’t need to plug in for that value because it will cancel out as we go along. So, this is our application of the Hall voltage equation for our first set of data. Now, let’s write a similar equation for our second set, again writing in all the information we have.

In this case, we write 3.0 microvolts is equal to 2.1 amps times 𝐵, which we don’t know and want to solve for, times 𝑑, which we don’t know, all divided by 𝑛 times 𝑞 times 𝐴, which we also don’t know for sure. All these variables we don’t know for sure may seem to be an issue. But if we divide one of these equations by the other, then that issue goes away. Because, thanks to this division, their significant cancellation of terms, the 𝑑, the 𝑛, the 𝑞, and the 𝐴, all drop out of this expression.

Looking at the expression that remains, we see that some more cancellation happens in terms of units. The microvolts units on the left-hand side cancels out and the amps unit on the right-hand side cancels. We have this remaining expression, which we then want to rearrange to solve for the magnetic field strength 𝐵. If we multiply both sides of the equation by 𝐵 and then multiply both sides of the equation by 3.0 divided by 3.5, then what we find is, on the right-hand side, the term 𝐵 cancels out. And on the left-hand side, the fraction 3.5 divided by 3.0 cancels with 3.0 divided by 3.5.

In the end, we find this expression for the unknown magnetic field 𝐵. And we’re ready to calculate this field. When we do, we find that, to two significant figures, it’s one 1.3 tesla. That’s the magnetic field that fits in with a Hall voltage of 3.0 microvolts in a current through our sample of 2.1 amps.