A solenoid that has length 𝐿 and 𝑁 turns is connected to a battery that has emf V. What would happen to the magnetic flux density at its axis if we compress the solenoid to decrease its length to half? (A) The magnetic flux density would halve. (B) The magnetic flux density would double. (C) The magnetic flux density would not change. (D) There is not enough information to determine the answer.
In this question, we’re being asked about a solenoid. And we can recall that a solenoid is a wire formed into a shape like this one, consisting of a series of equally spaced loops or turns. We’re told that this particular solenoid has a length of 𝐿. So that’s the length between these two ends of the solenoid. We’re also told that it consists of 𝑁 turns. So that’s 𝑁 of these individual loops or turns of the wire.
In the sketch that we’ve drawn, we can see that there are four of these loops, and so we would have 𝑁 is equal to four. It’s worth being clear though that this is just a sketch to get an idea of what’s going on. And in reality we don’t know the value of this quantity 𝑁. The other bit of information that we’re given is that this solenoid is connected to a battery that provides an emf of V. As a result of that emf, there must be a current in the wire. And let’s label that current as 𝐼.
We’re being asked about the magnetic flux density, also known as the strength of the magnetic field, on the axis of the solenoid. This axis is a line that passes through the center of the solenoid, as we’ve shown here in blue. Now, it turns out that there’s an equation for the magnetic flux density anywhere inside of a solenoid in terms of the current 𝐼 in the wire, the number of turns 𝑁 of that wire, and the solenoid’s length 𝐿. Specifically, we can recall that the magnetic flux density 𝐵 is equal to 𝜇 naught multiplied by 𝑁 multiplied by 𝐼 divided by 𝐿, where 𝜇 naught is a constant known as the permeability of free space.
So then, for the solenoid from this question with 𝑁 turns of wire, a length of 𝐿, and some current 𝐼 in that wire, the magnetic flux density inside of that solenoid has some value that we’ll call 𝐵 one equal to 𝜇 naught times 𝑁 times 𝐼 divided by 𝐿. We’re then told that the solenoid is compressed to half of its original length. That is, after it’s compressed, the new length of the solenoid is 𝐿 over two. It’s important to realize that nothing else about the solenoid has changed. It still consists of the same number of turns of wire 𝑁, just now distributed over this smaller length, 𝐿 over two.
It’s also still the same piece of wire connected to the same battery providing the same emf of V. That means that the current will still have the same value of 𝐼 as it did before compressing the solenoid. Let’s label the magnetic flux density inside the compressed solenoid as 𝐵 two. We know that this is equal to 𝜇 naught multiplied by the number of turns multiplied by the current divided by the solenoid’s length. In this case, that number of turns is 𝑁, the current is 𝐼, and the length is 𝐿 over two. And so we have that 𝐵 two is equal to 𝜇 naught multiplied by 𝑁 multiplied by 𝐼 divided by 𝐿 over two.
We can rewrite this expression as two times 𝜇 naught times 𝑁 times 𝐼 divided by 𝐿. If we then add some parentheses like this, we can notice that the terms inside those parentheses — so that’s 𝜇 naught times 𝑁 times 𝐼 over 𝐿 — are just equal to the magnetic flux density 𝐵 one inside of the original solenoid before compression. That is, we can take all these terms inside the parentheses and replace them with 𝐵 one. When we do that, we find that 𝐵 two is equal to two multiplied by 𝐵 one.
What this equation here is telling us is that if we compress a solenoid to half its original length, then the magnetic flux density 𝐵 two inside of the compressed solenoid is equal to twice the magnetic flux density 𝐵 one inside of the original uncompressed solenoid. That is, when we compress a solenoid to half its original length while keeping the number of turns and the current the same, the magnetic flux density inside that solenoid doubles. This conclusion matches the statement given in answer option (B).
We therefore choose option (B) as our answer. If we were to compress a solenoid to decrease its length to half, then the magnetic flux density would double.