Question

A solenoidal coil with 26 turns of wire is wound tightly around another coil with 350 turns. The inner solenoid is 20.0 cm long and has a diameter of 2.00 cm . At a certain time, the current in the inner solenoid is 0.100 A and is increasing at a rate of 1800 A/s .

Part A
For this time, calculate the average magnetic flux through each turn of the inner solenoid.

Part B
For this time, calculate the mutual inductance of the two solenoids;

Part C
For this time, calculate the emf induced in the outer solenoid by the changing current in the inner solenoid.

Answer 1

Part a)

`\phi = 2.76 * 10^(-7) T m^2`

Part B)

`M = 5.52 * 10^(-5) H`

Part C)

`EMF = 0.1 V/s`

Step-by-step explanation:

Part a)

Magnetic field due to a long ideal solenoid is given by

`B = \mu_0 n i`

n = number of turns per unit length

`n = (N)/(L)`

`n = (350)/(0.20)`

`n = 1750 turn/m`

now we know that magnetic field due to solenoid is

`B = (4\pi * 10^(-7))(1750)(0.100)`

`B = 2.2 * 10^(-4) T`

Now magnetic flux due to this magnetic field is given by

`\phi = B.A`

`\phi = (2.2 * 10^(-4))(\pi r^2)`

`\phi = (2.2 * 10^(-4))(\pi(0.02)^2)`

`\phi = 2.76 * 10^(-7) T m^2`

Part B)

Now for mutual inductance we know that

`\phi_(total) = M i`

`\phi_(total) = N\phi`

`\phi_(total) = 20(2.76 * 10^(-4))`

`\phi_(total) = 5.52 * 10^(-6)`

now we have

`M = (5.52 * 10^(-6))/(0.100)`

`M = 5.52 * 10^(-5) H`

Part C)

As we know that induced EMF is given as

`EMF = M (di)/(dt)`

`EMF = 5.52 * 10^(-5) (1800)`

`EMF = 0.1 V/s`