### Video Transcript

Consider the circular loops of the
current-carrying wire shown in this diagram. At the center of loop A, the
magnetic flux density is π΅ sub A. At the center of loop B, the
magnetic flux density is π΅ sub B. What is the value of π΅ sub A
divided by π΅ sub B? Is it (A) one, (B) two-thirds, (C)
three-halves, or (D) six?

To answer this question, we should
recall the formula for the magnetic flux density, or magnetic field strength, π΅, at
the center of a single current-carrying loop of wire. π΅ equals π naught πΌ divided by
two π, where π naught is a constant called the permeability of free space. πΌ is the current present in the
wire. And π is the radius of the
loop. Using the information given in the
diagram, we can apply this formula to loop A. π΅ sub A equals π naught times πΌ
over two times π. Likewise, for loop B, we have that
π΅ sub B equals π naught times three πΌ over two times two π.

Now, weβve been asked to find the
ratio of the magnetic flux density at the center of loop A to that of loop B. So we have this expression: π΅ A
over π΅ B. One simple way we can approach this
is to just substitute in each of these two equations for π΅. Doing this, we end up with a big
bulky expression, a fraction divided by a fraction. So letβs do some simplifying. Notice that in both the numerator
and denominator, we can group these terms together: π naught, πΌ, two, and π. These all came from the original
formula for π΅. So it makes sense that π΅ A and π΅
B have these in common.

Now, these cancel each other out of
the expression entirely. So all weβre left with is one over
three-halves. Remember that dividing by a
fraction is equivalent to multiplying by its reciprocal. So one over three-halves just
simplifies to two-thirds. This corresponds to answer option
(B), so we have our final answer. Weβve found that the ratio of the
magnetic flux density at the center of loop A to that of loop B equals
two-thirds.