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

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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΅_A. At the center of loop B, the magnetic flux density is π΅_B. What is the value of π΅_(A)/π΅_(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 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 πΌ divided by two π, where π naught is a constant called the permeability of free space. πΌ 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 πΌ over two times π. Likewise, for loop B, we have that π΅ sub B equals π naught times three πΌ over two times two π.

Now, weβ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: 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 letβs do some simplifying. Notice that in both the numerator and denominator, we can group these terms together: π naught, πΌ, 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 weβ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. Weβ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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