1. Coil 1 has 111 turns and coil 2 has 222 turns. When a constant current of 1.0

Question

1. Coil 1 has 111 turns and coil 2 has 222 turns. When a constant current of 1.010 A flows in coil 1 and no current flows in coil 2, the average magnetic flux from this 1.010 A current is 0.600 mWb through each turn of coil 1 and 0.200 mWb through each turn of coil 2. (Of course, the zero current in coil 2 gives zero magnetic flux through each turn of each coil.) Find the numerical value of the mutual inductance of the two coils.

Use the data in Problem 1 above to find the numerical value of the self-inductance of one of the two coils. (The self-inductance of the other coil is indeterminate.)

Explanation / Answer

As the current passing through coil 2 is zero hence all the flux through coil 2 would be coming from mutual inductance.

So 2 = M*I1   where 2 is the flux through coil 2, M is the mutual inductance

and I1 is the current through coil 1.

Now we are given: 2 = 0.2*222 mWb = 4.44*10-2 Wb; and I1 = 1.01 A

Hence M = 2/I1 = 4.44*10-2/1.01 = 4.396*10-2 Wb/A

Now we also have for coil 1:   1  = L1*I1   where L1 is the self inductance of coil 1.

So L1  = 0.6*111*10-3/1.01 = 0.066 Wb/A

Moreover the self inductance of coil 2 cannot be determined as no current is flowing through it.

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