Question

You have a 1.8m long copper wire. You want to make an N-turn current loop that generates a 2.0mT magnetic field at the center when the current is 1.3A . You must use the entire wire.. What will be the diameter of your coil?

Answer 1

Answer:

The diameter is  0.022m.

Explanation:

The magnetic field $B$ at the center of the coil is given by

(1). $B = \dfrac{\mu_0 NI}{d}$

where $\mu_0 = 1.26*10^{-6} m\:kg\:s^{-2} A^{-2}$ is the magnetic constant, $I$ is the current, $N$ number of coils, and $d$ is the diameter of the coil.

Now, if we call $L$ the length of the wire, then it must be true that

$\pi dN = L$ (this says $N$ coil circumferences ($c=\pi d$) fit into $L$ )

$\therefore N = \dfrac{L}{\pi d }$

putting this into equation (1) we get:

$B = \dfrac{\mu_0 IL }{\pi d^2}$

solve for $d:$

$\boxed{d = \sqrt{\dfrac{\mu_0I L }{\pi B} }}$

putting in the numerical values

$\mu_0 = 1.26*10^{-6} m\:kg\:s^{-2} A^{-2}$

$I =1.3A$

$L= 1.8m$

$B =2.0*10^{-3}T$

we get:

$d = \sqrt{\dfrac{(1.26*10^{-6})(1.3) *1.8 }{\pi (2.0*10^{-3})} }$

$\boxed{d = 0.022m}$