### Video Transcript

A circular coil of diameter 12
centimeters carries a current πΌ that generates a magnetic field in its center of
magnitude two times 10 to the power of negative four teslas. Given that π naught equals four π
times 10 to the power of negative seven henries per meter, calculate the magnetic
dipole moment of this coil. (A) 1.782 ampere meters squared,
(B) 3.6 ampere meters squared, (C) 0.0624 ampere meters squared, (D) 0.216 ampere
meters squared.

In this question, we are being
asked to calculate the magnetic dipole moment of this coil. We can recall that if we have a
loop of current-carrying wire, where the current has a magnitude πΌ and the area of
that loop is called π΄, then the magnetic dipole moment of the loop π sub π is
equal to πΌ times π΄. We are not given a value for the
current πΌ, so weβll have to use an equation substitution that uses values we are
given in the question.

We are told in the question that
the circular coil generates a magnetic field in its center. Now, we can recall that the
magnetic field strength π΅ at the center of a current-carrying loop of wire is given
by the equation π΅ equals π naught πΌ over two π, where πΌ is the current in the
loop, π is the radius of the loop, and π naught is the permeability of free
space.

Now, we can rearrange this equation
to make the current πΌ the subject. We can multiply both sides of the
equation by two π, and then divide both sides by π naught, to leave us with the
equation πΌ equals two ππ΅ over π naught. Substituting this into our equation
for π sub π, we have π sub π equals two ππ΅ over π naught multiplied by
π΄.

We can now find an expression for
the area π΄. The area of the circular coil is
the area of a circle, which is equal to ππ squared, where π is the radius of the
circle. Substituting this into our equation
for π sub π, we have π sub π equals two ππ΅ over π naught multiplied by ππ
squared. Alright then, we have our final
equation. So, letβs write out what values we
have for each variable.

The ones we are explicitly given
are the magnetic field strength, π΅, as two times 10 to the power of negative four
teslas and the permeability of free space, π naught, as four π times 10 to the
power of negative seven henries per meter. Then, we are given the diameter of
the loop of wire, 12 centimeters, but what we need is its radius, π. Before going further, letβs put
this value of diameter into the SI units of meters. We do this by dividing 12
centimeters by 100, as there are 100 centimeters in one meter. This gives us a value of 0.12
meters for the diameter.

Now then, the radius of the loop is
just half of its diameter. So, the radius is equal to 0.06
meters. Finally, with all of our variables
known, we can substitute them into the equation for the magnetic dipole moment as
follows. The units of teslas multiplied by
meters squared over henries gives us units of amperes. So, our final answer will be in
ampere meters squared. Solving the equation in our
calculators then, we get a value of 0.216 ampere meters squared for the magnetic
dipole moment of this coil.

If we now look at our answer
options, we can see that this answer corresponds with option (D). So, we can eliminate options (A),
(B), and (C) and confirm that option (D), 0.216 ampere meters squared, is the
correct answer for the magnetic dipole moment of this coil.