Φ = BA
The flux over the coil, equals the flux over one turn multiplied by the number of turns:
Φc = NΦ
Φc = NBA
The area of the coil is given by A = πR2, hence
Φc = πNBR2
"\u03a6c = \u03c0(70.0)(2.0 \\times 10^{-3}\\;T)(10 \\times 10^{-2} \\;m)^2 = 4.4 \\times 10^{-3} \\;Wb"
The net magnetic flux through the coil is zero, so the magnetic flux produced by the coil is given by equation:
"L = \\frac{\u03a6_c}{I}"
"L = \\frac{4.4 \\times 10^{-3} \\;Wb}{2.2 \\;A} = 2 \\times 10^{-3} \\;H"
Answer: 2 mH
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