### 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 sub air, is four π times 10 to the negative seven webers
per ampere meter, calculate the magnitude of the magnetic flux density at point
π. 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 π, 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 π, 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.

Letβs start with wire A. First, 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.

We know the current in wire A,
weβll call it πΌ sub A, is three amperes. The perpendicular distance from
wire A to the point π is 15 plus 10, or 25, centimeters. Weβll label this π sub A. Dividing this number in centimeters
by 100 to get a distance in meters, we find that π sub A equals 0.25 meters. Substituting these values, along
with the given value for π sub air, into our equation, we find the magnetic field
strength at point π 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 π 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 whatβs 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 πΌ
sub A. And so our fingers curl closed in
this direction. This means the magnetic field
direction at point π due to this current points out of the screen.

Now that we know the magnitude and
direction of the field at point π due to πΌ sub A, letβs calculate the field there
created by the current in wire B. Clearing some space to work, letβs
call the current in wire B πΌ sub B. We know it is two amperes. The perpendicular distance from
wire B to point π is 10 centimeters, which weβll label π sub B. Dividing this distance by 100, we
see that π 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 π due to
the current in wire B is 4.0 times 10 to the negative six teslas.

Once again, weβll use the
right-hand grip rule to find out the direction of the field created at point π,
this time due to the current πΌ 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 π 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 π. Clearing space at the top of our
screen, we add together π΅ sub A and π΅ 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 π.