Video: Magnetic Field Produced by a Long Straight Wire

A long straight horizontal wire carries a left-to-right current of 20 A. If the wire is placed in a uniform magnetic field of magnitude 4.0 × 10⁻⁵ T that is directed vertically downward, what is the resultant magnitude of the magnetic field 20 cm above the wire?

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Video Transcript

A long straight horizontal wire carries a left-to-right current of 20 amps. If the wire is placed in a uniform magnetic field of magnitude 4.0 times 10 to the negative fifth tesla that is directed vertically downward, what is the resultant magnitude of the magnetic field 20 centimeters above the wire?

So this example is interesting. We have a current-carrying wire, where the current we’re told is moving from left to right. And we’ll call this current capital 𝐼. In addition to that, this current-carrying wire is surrounded by a uniform magnetic field that we’re told points downward. And we can call that uniform field 𝐵 sub 𝑒. In this environment, we want to solve for the magnetic field magnitude at a position that’s 20 centimeters above this wire. Now, keep in mind that this wire is described as a long straight wire. And we’ll take that description to mean that it’s effectively infinite going off in either direction to the left and right.

So, that means it doesn’t matter much where along the wire we put our point where we want to solve for the magnetic field, just that it’s a distance 20 centimeters above the current-carrying wire. At this position, there are two magnetic fields which combine to form a net or total magnetic field. First, there’s the field due to the external magnetic field, what we’ve called 𝐵 sub 𝑒. That points, as we know, in a downward direction. But, secondly, there is the magnetic field created by the current in this wire.

And if we use our right-hand rule, with our thumb pointing in the direction of the current and our fingers curling in the direction of the field that’s created, then we know that the direction of the magnetic field created by this current 𝐼 is in circular loops around the wire that points out of the page above the wire and into the page below it.

So, if we were to draw that magnetic field acting at our point 𝑥 20 centimeters above the wire, we would show it pointing out of the page. That might look like this. And we can call this 𝐵 sub 𝑤, for the magnetic field created by the wire. These two magnetic fields acting at our point of interest are perpendicular to one another. And we want to solve for the magnitude of the resultant vector of these two components.

Since 𝐵 sub 𝑤 and 𝐵 sub 𝑒 are at right angles one to another, we can solve for this magnitude of the resultant vector this way. We can say that that overall magnitude, which we’ll call 𝐵, is equal to the square root of 𝐵 sub 𝑒 squared, the external magnetic field, plus 𝐵 sub 𝑤 squared, where 𝐵 sub 𝑤 is the magnetic field due to the wire. So then, to solve for 𝐵, we need to know the external field and the field from the wire. Now, we know the external field. That’s given to us in the problem statement. So, the one missing piece is the field created by our current-carrying wire.

To see what that field is, we can recall an equation describing the magnetic field created by a long straight current-carrying wire. That field is equal to 𝜇 nought, the permeability of free space, a constant, times the current in the wire divided by two 𝜋 times 𝑟, where 𝑟 is the perpendicular distance between the point of interest, where we wanna calculate the field and the axis of the current-carrying wire.

We can treat the free space permeability as 1.26 times 10 to the negative sixth tesla meters per ampere. And in addition to that, we know that 𝑟, our distance between the wire and our point of interest, is 20 centimeters, or converted to meters 0.20 meters. Since we also know the current 𝐼, as well as the external magnetic field 𝐵 sub 𝑒, we’re ready to plug in and then solve for 𝐵, the magnitude of the resulting magnetic field.

With all these values plugged in, notice briefly what happens to the units in the second term in our square root expression. The units of amperes cancel out, as do the units of meters, leaving us with the units of tesla. Our final calculated value for the magnetic field 𝐵 then will have those units.

And that value, to two significant figures, is 4.5 times 10 to the negative fifth tesla. That’s the magnetic field magnitude that results from adding the external magnetic field to the magnetic field created by the wire at the point we’ve chosen.

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