Answer on Electric Circuits Question for Herminda Bolando
N1 = 300
Number of turns in outer solenoid is
N2 = 30
Length of inner solenoid is
L = 21.0[cm] = 0.21[m]
Radius of inner solenoid is
r = diameter/2 = 2.00[cm]/2 = 1.00[cm] = 0.01[m]
The current in the inner solenoid is
i = 0.150[A]
Rate of increasing of current is
di/dt = 1800[A/s].
The average magnetic flux through each turn of the inner solenoid is
B.A = μ*(N/L)*i.A = [4(pi)*10^-7]*[300/0.21]*[0.15]*π*(0.01) = 8.45966*10^(-8) weber.
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