Magnetic flux through smaller coil within solenoid

[jdubt]

Homework Statement

A solenoid has 235 turns of wire wrapped along its length of 27 cm. The diameter of the solenoid is 1.5 cm. In the middle of the solenoid is a smaller coil of diameter 1.0 cm with 50 turns of wire along its 2.0 cm length. The two coils are coaxial. A current of 15 A is in the larger solenoid with no current in the smaller coil.
(a) Determine the magnetic flux through the smaller coil.
(b) Determine the mutual inductance of the pair of coils.

Homework Equations

B= mu_not *n*I
flux =B x A

The Attempt at a Solution

I've tried several times with no success. My problem is I don't know how to relate the solenoid to the smaller coil.

B= 4* pi *10^-7*(235/.27)*15 = .01641 T

I believe this is correct but I don't know where to go from here.

flux = (.01641 T)* pi * (.0075)^2

This could be an intermediary step in getting the flux of the smaller coil, but I don't know what to afterwards.

flux = (.01641 T)* pi * (.005)^2

This is incorrect.

Any help would be much appreciated.

 

[Redbelly98]

I agree with your 0.01641 T calculation.

Hmmm ... your flux expression seems right for a single-loop coil, but they give 50 turns for the smaller coil.

 

[baba89]

To determine the magnetic flux through the smaller coil within the solenoid, we can use the equation B = μ₀nI, where B is the magnetic field, μ₀ is the permeability of free space, n is the number of turns per unit length, and I is the current. Since the two coils are coaxial, we can assume that the magnetic field generated by the larger solenoid is uniform and parallel to the smaller coil.

Using the given values, we can calculate the magnetic field at the center of the smaller coil: B = (4π×10^-7 Tm/A)(235 turns/0.27 m)(15 A) = 0.01641 T. This value is the same as the one you calculated for the larger solenoid, indicating that the magnetic field is indeed uniform and parallel to the smaller coil.

Now, to calculate the magnetic flux through the smaller coil, we can use the equation flux = B × A, where A is the area of the coil. Since the smaller coil has a diameter of 1.0 cm, its radius is 0.5 cm or 0.005 m. Therefore, the area of the coil is A = πr² = π(0.005 m)² = 7.85×10^-5 m².

Plugging in the values, we get the flux through the smaller coil as: flux = (0.01641 T)(7.85×10^-5 m²) = 1.288×10^-6 Wb.

To determine the mutual inductance of the pair of coils, we can use the equation M = N₁N₂Φ/I, where N₁ and N₂ are the number of turns in the two coils, Φ is the magnetic flux through the smaller coil, and I is the current in the larger solenoid.

Plugging in the values, we get the mutual inductance as: M = (235 turns)(50 turns)(1.288×10^-6 Wb)/(15 A) = 1.614×10^-4 H. This tells us that there is a strong mutual inductance between the two coils, indicating that there will be a significant induced current in the smaller coil when the current in the larger solenoid changes.