### Video Transcript

A long straight wire is placed
perpendicular to a magnetic field of flux density five times 10 to the negative six
tesla, as shown in the figure below. The intensity of the current
passing through the wire is three amperes. Calculate the resultant magnetic
flux density at point 𝑃, given that 𝜇 air equals four 𝜋 times 10 to the negative
seven webers per ampere meter. (A) 5.8 times 10 to the negative
six tesla. (B) Eight times 10 to the negative
six tesla. (C) Five times 10 to the negative
six tesla. (D) Three times 10 to the negative
six tesla.

In this question, we are shown a
figure that shows a wire carrying a current of three amperes placed perpendicularly
to a magnetic field of strength five times 10 to the negative six tesla. We are asked to look at point 𝑃
and figure out what the resultant magnetic flux density is at that point, given that
the magnetic permeability of air, 𝜇 air, is four 𝜋 times 10 to the negative seven
webers per ampere meter.

In order to answer this question,
we’ll need to remind ourselves what happens when a current-carrying wire is placed
in a magnetic field. Recall that when a wire is carrying
a current, it produces a magnetic field that is directed around the wire. The direction of the field around
the wire is given by the right-hand rule. This rule tells us that if we point
the thumb of our right hand along the current direction with our fingers
outstretched, then when we curl our fingers around, the direction that our fingers
curl is the direction of the magnetic field.

In this case, since the current is
directed upward, then the right-hand rule tells us that the magnetic field due to
this current, which we’ve called 𝐵 sub 𝐼, is directed as we have indicated on the
diagram. That is, the magnetic field due to
the current-carrying wire is directed into the screen on the right-hand side of the
wire and out of the screen on the left-hand side.

On the right-hand side, where the
point 𝑃 that we’re interested in is, we see that this is the same direction as the
existing magnetic field that the wire is placed in. Since the two fields are in the
same direction at point 𝑃, that means that their magnitudes will simply add
together to give the resultant field strength at this point.

Calling the background field
strength 𝐵 sub 𝑏 and the field due to the wire at point 𝑃, 𝐵 sub 𝐼 𝑃, we have
that the resultant field strength or flux density at point 𝑃, 𝐵 sub 𝑇, is equal
to 𝐵 sub 𝐼 𝑃 plus 𝐵 sub 𝑏. We know the value of the background
flux density, 𝐵 sub 𝑏, so we need to calculate the value of 𝐵 sub 𝐼 𝑃, the flux
density at point 𝑃 due to the current, and add the two values together.

Recall that the magnetic field due
to a current-carrying wire, 𝐵 sub 𝐼, can be described using the equation 𝐵 sub 𝐼
is equal to 𝜇 air, the permeability of air, multiplied by 𝐼, the current in the
wire, all divided by two 𝜋 times the distance 𝑟 away from the wire. Notice that the distance from the
wire is in the denominator, which means that the strength of the magnetic field will
decrease as distance from the wire increases.

Let’s now take a look at the values
of the relevant quantities at point 𝑃, in order to figure out what the magnetic
field produced by the wire is at this point. At point 𝑃, the distance away from
the wire is 20 centimeters. This is our value for 𝑟. We also know that the current, 𝐼,
in the wire is three amperes. Finally, we’ve been given a value
of four 𝜋 times 10 to the negative seven webers per ampere meter for 𝜇 air.

Before using these values to
calculate the magnetic flux density 𝐵 sub 𝐼 𝑃, we need to convert the distance
from centimeters into meters. Remember that there are 100
centimeters in one meter. So to convert this value of 𝑟 from
centimeters to meters, we need to multiply it by 10 to the negative two meters per
centimeter. This works out as 0.2 meters. Let’s now substitute our values for
the current, 𝐼, the distance, 𝑟, and the permeability, 𝜇 air, into this equation
for the magnetic flux density.

We have that 𝐵 sub 𝐼 𝑃 is equal
to four 𝜋 times 10 to the negative seven webers per ampere meter multiplied by
three amperes divided by two 𝜋 times 0.2 meters. Let’s now have a look at the units
in the expression. In the numerator of the right-hand
side, we have webers per ampere meter multiplied by amperes. Then in the denominator, we just
have units of meters. Grouping these units together, we
have units of weber amperes per ampere meter squared. We can cancel the amperes from the
numerator and denominator, leaving units of webers per meter squared. Now, we might recall that units of
webers per meter squared are equal to units of tesla, the SI unit of magnetic flux
density.

Now we know that our units are
correct, we can go ahead and evaluate this expression. We find that the magnetic flux
density at point 𝑃 due to the current-carrying wire is equal to three times 10 to
the negative six tesla. Remember though that this value is
only the magnetic flux density due to the wire. We still need to add to this the
magnetic flux density due to the constant magnetic field that the wire is in.

Adding together the field due to
the current and the background field, we have that the total field strength at point
𝑃 is equal to three times 10 to the negative six tesla plus five times 10 to the
negative six tesla. This gives us a value of eight
times 10 to the negative six tesla. This matches the value given in
option (B). So our answer then is option
(B). The resultant magnetic flux density
at point 𝑃 is eight times 10 to the negative six tesla.