Question

You have a 3-m-long copper wire. You want to make an N-turn current loop that generates a 2.795 mT magnetic field at the center when the current is 1.29 A. You must use the entire wire. What will be the diameter, in cm, of your coil?

Answer 1

Answer tab

First, we need to remember the magnetic field in the center of a coil, which is:

`B=(\mu_0iN)/(2R)`

Where R is the radius, and N is the number of turns. So, we have two unknows. However, the total length of the wire also gives us a constraint to the problem. We can equate this as: (L is the total length of 3m)

`L=(2\pi R)N`

As 2piR is the length of a single turn, multiplied by the number of turns. If we replace our values on these equations, we get:

`2.795*10^(-3)=(\mu_0*1.29*N)/(2R)`

And

`3=2\pi NR`

So, we have two equations and two unknowns. We can find out their values. As we want to find the diameter, let us first isolate the value of N on the second equation:

`N=(3)/(2\pi R)`

By replacing this value of N on the first equation, we get:

`2.795*10^(-3)=(\mu_0*1.29)/(2R)*(3)/(2\pi R)`

Then, if we isolate R, we get:

`R^2=(\mu_0*1.29*3)/(4\pi *2.795*10^(-3))`

Then our final answer is D=2R=2.35cm