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 𝐵 sub A. At the center of loop B, the
magnetic flux density is 𝐵 sub B. What is the value of 𝐵 sub A
divided by 𝐵 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, 𝐵, at
the center of a single current-carrying loop of wire. 𝐵 equals 𝜇 naught 𝐼 divided by
two 𝑟, where 𝜇 naught is a constant called the permeability of free space. 𝐼 is the current present in the
wire. And 𝑟 is the radius of the
loop. Using the information given in the
diagram, we can apply this formula to loop A. 𝐵 sub A equals 𝜇 naught times 𝐼
over two times 𝑟. Likewise, for loop B, we have that
𝐵 sub B equals 𝜇 naught times three 𝐼 over two times two 𝑟.
Now, we’ve 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: 𝐵 A
over 𝐵 B. One simple way we can approach this
is to just substitute in each of these two equations for 𝐵. Doing this, we end up with a big
bulky expression, a fraction divided by a fraction. So let’s do some simplifying. Notice that in both the numerator
and denominator, we can group these terms together: 𝜇 naught, 𝐼, two, and 𝑟. These all came from the original
formula for 𝐵. So it makes sense that 𝐵 A and 𝐵
B have these in common.
Now, these cancel each other out of
the expression entirely. So all we’re 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. We’ve found that the ratio of the
magnetic flux density at the center of loop A to that of loop B equals
two-thirds.