Problem 3:
A circular coil of wire (radius = 0.5 m, 12 turns) is located in the xy plane. The total resistance of the coil is 5
øB = N
B ·
ñ dA
If the coil is in the x-y plane, then the normal to the surface ñ
is in the k or -k direction. Thus the dot product is
easy ->
B · ñ = (0.5i + 0.7k) ·
k
= 0.7 T (we are only concerned with the magnitude of the flux, so we don't need to worry about the possible minus sign). This is constant over the area of
the coil, so the integral is just the total area =
(0.5 m)2 and the
total flux is
øB = (12)(0.7 T)( (0.5 m)2) = 6.60 Wb
E = - døB/dt = - d/dt [1-0.4t](N r2) = 0.4 N
r2 = 0.4 (12) (
(0.5 m)2) = 3.77 V
Using V = IR, the current in the coil will be
I = E/R = 3.77V/5 = 0.754 A.
Problem 4
Test 3
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