Answer:
radius of the loop = 7.9 mm
number of turns N ≅ 399 turns
Explanation:
length of wire L= 2 m
field strength B = 3 mT = 0.003 T
current I = 12 A
recall that field strength B = μnI
where n is the turn per unit length
vacuum permeability μ = [tex]4\pi *10^{-7} T-m/A[/tex] = 1.256 x 10^-6 T-m/A
imputing values, we have
0.003 = 1.256 x 10^−6 x n x 12
0.003 = 1.507 x 10^-5 x n
n = 199.07 turns per unit length
for a length of 2 m,
number of loop N = 2 x 199.07 = 398.14 ≅ 399 turns
since there are approximately 399 turns formed by the 2 m length of wire, it means that each loop is formed by 2/399 = 0.005 m of the wire.
this length is also equal to the circumference of each loop
the circumference of each loop = [tex]2\pi r[/tex]
0.005 = 2 x 3.142 x r
r = 0.005/6.284 = [tex]7.9*10^{-4} m[/tex] = 0.0079 m = 7.9 mm
