Answer:
The diameter is 0.022m.
Explanation:
The magnetic field [tex]B[/tex] at the center of the coil is given by
(1). [tex]B = \dfrac{\mu_0 NI}{d}[/tex]
where [tex]\mu_0 = 1.26*10^{-6} m\:kg\:s^{-2} A^{-2}[/tex] is the magnetic constant, [tex]I[/tex] is the current, [tex]N[/tex] number of coils, and [tex]d[/tex] is the diameter of the coil.
Now, if we call [tex]L[/tex] the length of the wire, then it must be true that
[tex]\pi dN = L[/tex] (this says [tex]N[/tex] coil circumferences ([tex]c=\pi d[/tex]) fit into [tex]L[/tex] )
[tex]\therefore N = \dfrac{L}{\pi d }[/tex]
putting this into equation (1) we get:
[tex]B = \dfrac{\mu_0 IL }{\pi d^2}[/tex]
solve for [tex]d:[/tex]
[tex]\boxed{d = \sqrt{\dfrac{\mu_0I L }{\pi B} }}[/tex]
putting in the numerical values
[tex]\mu_0 = 1.26*10^{-6} m\:kg\:s^{-2} A^{-2}[/tex]
[tex]I =1.3A[/tex]
[tex]L= 1.8m[/tex]
[tex]B =2.0*10^{-3}T[/tex]
we get:
[tex]d = \sqrt{\dfrac{(1.26*10^{-6})(1.3) *1.8 }{\pi (2.0*10^{-3})} }[/tex]
[tex]\boxed{d = 0.022m}[/tex]
