A solenoidal coil with 30 turns of wire is wound tightly around another coil with 340 turns. The inner solenoid is 21.0 long and has a diameter of 2.50 . At a certain time, the current in the inner solenoid is 0.110 and is increasing at a rate of 1800 . For this time, calculate the average magnetic flux through each turn of the inner solenoid.
Number of turns in inner solenoid is
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
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