Question Video: Finding the Ratio of the Magnetic Flux Densities at the Centers of Two Circular Loops of Wire | Nagwa Question Video: Finding the Ratio of the Magnetic Flux Densities at the Centers of Two Circular Loops of Wire | Nagwa

Question Video: Finding the Ratio of the Magnetic Flux Densities at the Centers of Two Circular Loops of Wire Physics • Third Year of Secondary School

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_A. At the center of loop B, the magnetic flux density is 𝐵_B. What is the value of 𝐵_(A)/𝐵_(B)?

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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.

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