Acircular coil has a 10.0 cm radius and consists of 30.0 closely wound turns of wire. an externally produced magnetic field of magnitude 2.60 mt is perpendicular to the coil. (a) if no current is in the coil, what magnetic flux links its turns? (b)when the current in the coil is 3.80 a in a certain direction, the net flux through the coil is found to vanish. what is the inductance of the coil?

Magnetic flux can be calculated by the product of the magnetic field and the area that is perpendicular to the field that it penetrates. It has units of Weber or Tesla-m^2. For the first question, when there is no current in the coil, the flux would be:

ΦB = BA
          A = πr^2
          A = π(.1 m)^2
          A = π/100 m^2   

 ΦB = 2.60x10^-3 T (π/100 m^2 ) ΦB = 8.17x10^-5 T-m^2 or Wb (This is only for one loop of the coil)

The inductance on the coil given the current flows in a certain direction can be calculated by the product of the total number of turns in the coil and the flux of one loop over the current passing through. We do as follows:

L = N (ΦB ) / I
L = 30 (8.17x10^-5 T-m^2) / 3.80 = 6.44x10^-4 mH

Letter a, because when there are unlike charges they will always attract to each other.