Question Video: Understanding the Link Between Resistivity and Magnetic Flux Density of a Solenoid Physics • Third Year of Secondary School

Video Transcript

Two wires have the same cross-sectional areas and lengths. One is made from copper and the other from aluminium. If the wires are reshaped into two identical solenoids and each is connected to an electric source of emf equal to 12 volts, which of the following will be true about their magnetic flux density, given that aluminium resistivity is higher than copper resistivity? (A) 𝐵 copper is equal to 𝐵 aluminium. (B) 𝐵 copper is greater than 𝐵 aluminium. (C) 𝐵 copper is less than 𝐵 aluminium. (D) The answer cannot be determined.

This question is asking us about two solenoids, one of which is made from copper and the other from aluminium. We can recall that a solenoid is a wire that’s formed into a series of equally spaced loops or turns, as in the diagrams that we’ve drawn here. In this case, we’re told that each of these two solenoids is made of wire that has the same cross-sectional area and the same length.

We’re also told that the solenoids themselves are identically shaped to each other. That is, each of the solenoids has the same length, which we’ve labeled here as 𝐿, and each consists of the same number of turns of wire, which we’ve labeled as 𝑁. We’re also told that each of these two solenoids is connected to an electric source providing the same emf of 12 volts. In fact, the only difference between the two solenoids is the material copper or aluminium that each is made of.

We’re told that the resistivity of aluminium, which we’ve labeled here as 𝜌 a, is greater than the resistivity of copper, 𝜌 c. We can recall that the resistivity 𝜌 of a wire is equal to its resistance 𝑅 multiplied by the wire’s cross-sectional area divided by the wire’s length.

Now, we’re told that these two wires have the same cross-sectional area and the same length as each other. That means that if the resistivity 𝜌 a is greater than the resistivity 𝜌 c, then it must also be true that the resistance of the aluminium wire 𝑅 a is greater than the resistance of the copper wire 𝑅 c. So we know then that the wire used to form this aluminium solenoid has a greater resistance than the wire used to form the copper solenoid.

Since each solenoid is connected to an electric source of emf, then we know that there’ll be a current in each of the wires. We’ve labeled this current as 𝐼 c for the copper wire and 𝐼 a for the aluminium wire. At this point, it’s gonna be helpful to recall Ohm’s law, which tells us that the applied emf 𝑉 is equal to the current 𝐼 multiplied by the resistance 𝑅. If we divide both sides of Ohm’s law by 𝑅 so that on the right-hand side the 𝑅’s cancel each other out, then we can rewrite this Ohm’s law equation as current 𝐼 is equal to emf 𝑉 divided by resistance 𝑅.

Now, for both these two solenoids, we know that the emf 𝑉 is equal to 12 volts. We have then that for the copper solenoid, the current 𝐼 c is equal to 12 volts divided by the resistance 𝑅 c. Similarly, for the aluminium one, we have 𝐼 a is equal to 12 volts divided by 𝑅 a. In both these two expressions for the currents 𝐼 c and 𝐼 a, the numerator on the right-hand side has the same value of 12 volts.

In terms of the denominators, we know that the resistance 𝑅 a is greater than the resistance 𝑅 c. That means that in this case for the aluminium solenoid, we’re dividing the emf of 12 volts by a larger resistance value than we are in the case of the copper solenoid. So, because the resistance of the aluminium wire is larger than the resistance of the copper wire, the current in the aluminium wire 𝐼 a will be smaller than the current 𝐼 c in the copper wire.

Now, the current in the wire of a solenoid creates a magnetic field inside of that solenoid. This magnetic field is what we’re being asked to compare between the two solenoids. Specifically, we’re being asked which of these answers is true about the magnetic flux density, also known as the strength of the magnetic field inside of each of the two solenoids.

The key result that we’ve reached so far is this one here, comparing the currents in each of the two solenoids. Let’s now clear some space on the screen to see how we can use this result to work out which of these statements is true.

We know that the current 𝐼 a in the aluminium wire is less than the current 𝐼 c in the copper wire. We’ve also already said that it’s this current in the wire that causes there to be a magnetic field inside of the solenoid. We can recall that there’s an equation that tells us how to find the magnetic flux density or the strength of the magnetic field inside of a solenoid in terms of the current in the wire, the number of turns of wire, and the solenoid length. Specifically, the magnetic flux density 𝐵 is equal to a constant 𝜇 naught, known as the permeability of free space, multiplied by the number of turns of wire 𝑁 multiplied by the current 𝐼 in that wire divided by the solenoid length 𝐿.

Now, we know that both these two solenoids have the same number of turns 𝑁 and the same length 𝐿. We also know that this constant 𝜇 naught must of course have the same value in each case. The only quantity on the right-hand side of this equation that’s different for each of the two solenoids in the question is the current 𝐼. We have then that the magnetic flux density 𝐵 copper inside of the copper solenoid is equal to 𝜇 naught 𝑁 over 𝐿 multiplied by the current 𝐼 c, while 𝐵 aluminium, the magnetic flux density inside the aluminium solenoid, is equal to the same 𝜇 naught over 𝐿 multiplied by the current 𝐼 a.

What’s important is that these quantities in the parentheses on the right-hand side of each equation have the same value for each of the two solenoids. This means that any difference between the magnetic flux densities 𝐵 copper and 𝐵 aluminium is due to the difference between the currents 𝐼 c and 𝐼 a. We see that for constant values of 𝜇 naught, 𝑁, and 𝐿, the larger the current 𝐼, the larger the magnetic flux density 𝐵 will be.

Now, we know that 𝐼 a, the current in the aluminium wire, is smaller than 𝐼 c, the current in the copper wire. That then means that the magnetic flux density 𝐵 aluminium must be smaller than the magnetic flux density 𝐵 copper. Of course, we can equally flip that statement around to say that 𝐵 copper is greater than 𝐵 aluminium. That is, the magnetic flux density inside the copper solenoid is greater than that inside the aluminium one.

Since we have derived an equality relating these two magnetic flux densities, that means that we can rule out answer option (D), which claims that the answer cannot be determined. If we now compare this equality that we’ve derived to the ones given in answer options (A), (B), and (C), we find that our equality matches the one given in option (B). That means that our answer is that the statement that’s true about the magnetic flux densities is the one in option (B). 𝐵 copper is greater than 𝐵 aluminium.