The portal has been deactivated. Please contact your portal admin.

Question Video: Comparing the Magnetic Fields at the Centers of Two Circular Loops of Wire Physics

A circular loop of wire of radius 50 mm carries a constant current 𝐼 A and produces a magnetic field of strength 𝐡₁ T at its center. Another circular loop of wire has a radius of 150 mm. Given that this wire also carries a constant current of 𝐼 A, which of the following correctly shows the relation between 𝐡₂, the strength of the magnetic field produced by the larger loop at its center, and 𝐡₁? [A] 𝐡₂ = 1/3 𝐡₁ [B] 𝐡₂ = 3 𝐡₁ [C] 𝐡₂ = 1/9 𝐡₁ [D] 𝐡₂ = 9 𝐡₁ [E] 𝐡₂ = 𝐡₁

03:33

Video Transcript

A circular loop of wire of radius 50 millimeters carries a constant current 𝐼 amperes and produces a magnetic field of strength 𝐡 one teslas at its center. Another circular loop of wire has a radius of 150 millimeters. Given that this wire also carries a constant current of 𝐼 amperes, which of the following correctly shows the relation between 𝐡 two, the strength of the magnetic field produced by the larger loop at its center, and 𝐡 one? Is it (A) 𝐡 two equals one-third 𝐡 one? (B) 𝐡 two equals three 𝐡 one. (C) 𝐡 two equals one-ninth 𝐡 one. (D) 𝐡 two equals nine 𝐡 one. Or (E) 𝐡 two equals 𝐡 one.

In this question, we have two different loops of wire, we’ll call them loops one and two, that are carrying the same current but have different radii. And we want to compare the strength of the magnetic field produced at the centers of the loops. We were told that a magnetic field of strength 𝐡 one is produced at the center of the loop with a radius of 50 millimeters. So let’s go ahead and label this radius value as π‘Ÿ one. Likewise, we know that a magnetic field of strength 𝐡 two is produced at the center of the loop with a radius of 150 millimeters. So let’s call this radius π‘Ÿ two.

To answer this question, we should recall the formula for calculating the strength of the magnetic field 𝐡 produced at the center of a single wire loop of radius π‘Ÿ that’s carrying a current 𝐼, that is, 𝐡 equals πœ‡ naught 𝐼 divided by two π‘Ÿ. Note that the term πœ‡ naught is a known constant called the permeability of free space. Since we want to determine the relationship between magnetic field strength and the radius of a loop of wire, a proportionality relation will be really useful to us here.

Recall that a statement of proportionality tells how variables change with respect to each other. So, to create a proportionality from this formula, we simply ignore all the constant, unchanging values by setting them equal to one. And we replace the equals sign with this symbol, which tells us that we’re no longer strictly equating the left and right sides of this expression. Notice too that we treat the current 𝐼 as a constant here, since we know its value is the same for both wire loops. Thus, our relationship reads that 𝐡 is proportional to one over π‘Ÿ.

Another way to say this is that the strength of the produced magnetic field is inversely proportional to the radius of the loop, since as one quantity increases, the other must decrease. This makes sense because the magnetic field strength decreases with distance from a current-carrying wire. Just knowing this, we can eliminate a few answer options.

Loop two has a greater radius than loop one. So we know that the strength of the field at the center of loop two must be less than that of loop one. Therefore, we can eliminate answer options (B) and (D), because they suggest that 𝐡 two is greater than 𝐡 one. We can also eliminate option (E), since it suggests that the strengths of the magnetic fields are the same. This leaves options (A) and (C). And here’s where that statement of proportionality really comes in handy. 𝐡 is inversely proportional to π‘Ÿ. And so an increase in π‘Ÿ corresponds to a decrease in 𝐡 of the same factor.

Compared to loop one, loop two has a radius that’s three times greater. Thus, due to this inversely proportional relationship, the field strength at the center of loop two will be three times smaller. Answer choice (A) shows this relationship. So we know this is the correct answer. 𝐡 two equals one-third times 𝐡 one.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.