### 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 π, given that π 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 π 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, weβll 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 π and then work
out the resultant field.

To start off, letβs recall the
formula for a magnetic field around a current-carrying wire. The magnetic field strength π΅ at a
point a distance π from the wire is equal to π naught πΌ divided by two ππ,
where π naught is the permeability of free space and πΌ is the current in the
wire. Clearing some space on screen,
letβs also make a list of the values and variables given to us so far. Since weβll 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.

Letβs call the magnetic field at
point π due to wire one π΅ one. Our equation for π΅ one looks like
this. πΌ one is two amperes. We will treat wire one as being
directly above point π so that π 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 π 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
πΌ one, we find that the field at point π from wire one points to the right.

Moving on to wire two, we can
perform a similar calculation. πΌ two is three amperes, π two is
15 centimeters, or 0.15 meters, and π naught keeps its constant value. We find that π΅ two, the strength
of the magnetic field at point π 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 π 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 canβt 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 π΅ one and one for π΅ two. The net magnetic flux density is
simply the resultant of these two vectors, which weβll call π΅ net. If we draw π΅ 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 π΅
net, we can use Pythagorasβs theorem: π΅ net squared equals π΅ one squared plus π΅
two squared. Or, to write this another way, π΅
net equals the square root of π΅ one squared plus π΅ two squared. Plugging in our values for π΅ one
and π΅ two, we see that the overall strength of the magnetic field at point π,
rounded to one decimal place, is 4.8 times 10 to the negative six teslas. So, this is our final answer to
this question.