Question #15971

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.

Expert's answer

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

B.A = μ*(N/L)*i.A = [4(pi)*10^-7]*[300/0.21]*[0.15]*π*(0.01) = 8.45966*10^(-8) weber.

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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