Answer:
0.04895m
Explanation:
Given magnetic field [tex]B=1.5mT=1.5\times 10^{-3}T[/tex], Length of wire [tex]l=2.2m[/tex], current [tex]I=1.3A[/tex] and [tex]\mu_o=4\pi .10^-^7 N.A^-^2[/tex], the diameter,[tex]d[/tex] of the coil is calculated as:
[tex]B=\frac{N\mu_o I}{d}, l=N2\pi r=N\pi d, N=l/\pi d\\\\\therefore d^2=\frac{l\mu_o I}{B} \\\\d=\sqrt{\frac{l\mu_o I}{B} }\\\\d=\sqrt{\frac{2.2\times 4\pi \times 10^{-7}\times 1.3}{0.0015}}\\\\d=0.04895m[/tex]
Hence, the diameter of the coil is 0.04895m
