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Question Video: Calculating the Net Magnetic Flux Density Induced by Two Wires Physics

Two straight wires, A and B, are carrying currents of intensities 3 A and 2 A, respectively, as shown in the figure below. Given that the magnetic permeability of air, πœ‡_air, is 4πœ‹ Γ— 10⁻⁷ Wb/Aβ‹…m, calculate the magnitude of the magnetic flux density at point 𝑋. Give your answer in scientific notation to one decimal place.

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Video Transcript

Two straight wires, A and B, are carrying currents of intensities three amperes and two amperes, respectively, as shown in the figure below. Given that the magnetic permeability of air, πœ‡ sub air, is four πœ‹ times 10 to the negative seven webers per ampere meter, calculate the magnitude of the magnetic flux density at point 𝑋. Give your answer in scientific notation to one decimal place.

This question is asking us to calculate the magnitude, or size, of the magnetic field due to the two current-carrying wires at point 𝑋, located to the left of wire B. Each of the current-carrying wires produces a magnetic field. Therefore, to find the net magnetic flux density at point 𝑋, we have to do two calculations: one for the magnetic field created by wire A and one for the magnetic field created by wire B.

Let’s start with wire A. First, let’s recall the formula for a magnetic field around a current-carrying wire. The magnetic field strength 𝐡 at a point a distance π‘Ÿ from the wire is equal to πœ‡ naught 𝐼 divided by two πœ‹π‘Ÿ, where πœ‡ naught is the permeability of free space and 𝐼 is the current in the wire.

We know the current in wire A, we’ll call it 𝐼 sub A, is three amperes. The perpendicular distance from wire A to the point 𝑋 is 15 plus 10, or 25, centimeters. We’ll label this π‘Ÿ sub A. Dividing this number in centimeters by 100 to get a distance in meters, we find that π‘Ÿ sub A equals 0.25 meters. Substituting these values, along with the given value for πœ‡ sub air, into our equation, we find the magnetic field strength at point 𝑋 due to the current in wire A is 2.4 times 10 to the negative six teslas.

Magnetic fields are vector quantities. This means the directions of the fields at point 𝑋 due to the currents in wires A and B will affect the overall magnetic flux density at that point.

The direction of the magnetic field due to a long straight current-carrying wire is determined using what’s called the right-hand grip rule. If we point our right thumb in the direction of current in the wire, our fingers curl closed in the direction of the field. Applying this rule to wire A, we point our right thumb toward the top of the screen, the direction of the current 𝐼 sub A. And so our fingers curl closed in this direction. This means the magnetic field direction at point 𝑋 due to this current points out of the screen.

Now that we know the magnitude and direction of the field at point 𝑋 due to 𝐼 sub A, let’s calculate the field there created by the current in wire B. Clearing some space to work, let’s call the current in wire B 𝐼 sub B. We know it is two amperes. The perpendicular distance from wire B to point 𝑋 is 10 centimeters, which we’ll label π‘Ÿ sub B. Dividing this distance by 100, we see that π‘Ÿ sub B is equal to 0.10 meters. Substituting these values into our equation for the magnetic field, we find that the field strength at point 𝑋 due to the current in wire B is 4.0 times 10 to the negative six teslas.

Once again, we’ll use the right-hand grip rule to find out the direction of the field created at point 𝑋, this time due to the current 𝐼 sub B. Just like before, our right thumb points upward in the direction of the current, and our fingers grip or curl closed like this. Therefore, the magnetic field at point 𝑋 due to this second current-carrying wire also points out of the screen.

Since the two fields both point in the same direction, we can simply add them to find the net magnetic flux density at point 𝑋. Clearing space at the top of our screen, we add together 𝐡 sub A and 𝐡 sub B. This is 2.4 times 10 to the negative six teslas plus 4.0 times 10 to the negative six teslas, which gives us an answer of 6.4 times 10 to the negative six teslas. This is the total magnetic flux density at point 𝑋.

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