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

Video Transcript

Consider two long straight current-carrying wires. One is carrying a current of intensity two amperes, and the other is carrying a current of intensity three amperes. The two wires are perpendicular, as shown in the figure below. Calculate the net magnetic flux density at point 𝑃, given that 𝜇 equals four 𝜋 times 10 to the negative seven webers per ampere meter. Give your answer in scientific notation to one decimal place.

This question is asking us to calculate the net magnetic flux density at point 𝑃 due to two perpendicular current-carrying wires. Wire one has a current of two amperes going out of the page. Wire two has a current of three amperes going to the right. We can arbitrarily define positive and negative directions in this picture. For our purposes, we’ll say that out of the page, up, and to the right are positive directions. To solve this problem, we need to separately calculate the magnetic fields due to each wire at point 𝑃 and then work out the resultant field.

To start off, let’s recall the formula for a magnetic field around a current-carrying wire. The magnetic field strength 𝐵 at a point a distance 𝑟 from the wire is equal to 𝜇 naught 𝐼 divided by two 𝜋𝑟, where 𝜇 naught is the permeability of free space and 𝐼 is the current in the wire. Clearing some space on screen, let’s also make a list of the values and variables given to us so far. Since we’ll be doing several calculations, this will help us keep our numbers straight. The quantities labeled with a subscript one refer to wire one. And the quantities labeled with a subscript two refer to wire two.

Let’s call the magnetic field at point 𝑃 due to wire one 𝐵 one. Our equation for 𝐵 one looks like this. 𝐼 one is two amperes. We will treat wire one as being directly above point 𝑃 so that 𝑟 one is 15 centimeters. We know further that the magnetic permeability is four 𝜋 times 10 to the negative seven webers per ampere meter. Substituting in this value and converting the distance in centimeters to one in meters, we find that the strength of the magnetic field at point 𝑃 due to wire one is 2.7 times 10 to the negative six teslas.

Using the right-hand-grip rule for magnetic fields, we can find the direction of the magnetic field. If we point our right thumb out of the page in the direction of the current and curl our fingers closed, that curl direction tells us how the magnetic field created by the current points. Applying this rule to the current 𝐼 one, we find that the field at point 𝑃 from wire one points to the right.

Moving on to wire two, we can perform a similar calculation. 𝐼 two is three amperes, 𝑟 two is 15 centimeters, or 0.15 meters, and 𝜇 naught keeps its constant value. We find that 𝐵 two, the strength of the magnetic field at point 𝑃 due to wire two, is 4.0 times 10 to the negative six teslas. Again, using the right-hand grip rule, we see that the magnetic field at point 𝑃 due to the current in wire two points out of the page.

Now that we have our two components of magnetic field strength, we have to combine them. It might be tempting to just add our two values together outright, but we can’t do this. Because the wires are perpendicular to each other, their magnetic field components point in different directions. We can represent this by drawing two perpendicular vectors, one for 𝐵 one and one for 𝐵 two. The net magnetic flux density is simply the resultant of these two vectors, which we’ll call 𝐵 net. If we draw 𝐵 net onto our diagram, we see that we have a right-angled triangle, where the length of each side is the magnitude of each vector.

So, to find the magnitude of 𝐵 net, we can use Pythagoras’s theorem: 𝐵 net squared equals 𝐵 one squared plus 𝐵 two squared. Or, to write this another way, 𝐵 net equals the square root of 𝐵 one squared plus 𝐵 two squared. Plugging in our values for 𝐵 one and 𝐵 two, we see that the overall strength of the magnetic field at point 𝑃, rounded to one decimal place, is 4.8 times 10 to the negative six teslas. So, this is our final answer to this question.