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?

Respuesta :

Answer tab

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

[tex]B=\frac{\mu_0iN}{2R}[/tex]

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)

[tex]L=(2\pi R)N[/tex]

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:

[tex]2.795*10^{-3}=\frac{\mu_0*1.29*N}{2R}[/tex]

And

[tex]3=2\pi NR[/tex]

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:

[tex]N=\frac{3}{2\pi R}[/tex]

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

[tex]2.795*10^{-3}=\frac{\mu_0*1.29}{2R}*\frac{3}{2\pi R}[/tex]

Then, if we isolate R, we get:

[tex]R^2=\frac{\mu_0*1.29*3}{4\pi *2.795*10^{-3}}[/tex]

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