You have two cylindrical solenoids, one inside the other, with the two cylinders concentric. The outer solenoid has length ℓouter = 400 mm, radius R outer = 50 mm, and N outer = 2000 windings. The inner solenoid has length ℓinner = 40 mm, radius R inner = 20 mm, and N inner = 150 windings. The inner solenoid is centered within the outer solenoid.
When the outer solenoid carries a current given by I(t)=I0sin(ωt), with I0 = 600 mA and ω = 100 s−1, what is the peak emf in the inner solenoid?
To find the peak emf in the inner solenoid, we can use Faraday's law of electromagnetic induction. This law states that the induced emf (ε) in a circuit is equal to the negative rate of change of magnetic flux through the circuit.
In this case, the inner solenoid is centered within the outer solenoid, and the outer solenoid is carrying a time-varying current. Therefore, the magnetic field produced by the outer solenoid will vary with time, and this changing magnetic field will induce an emf in the inner solenoid.
To calculate the peak emf in the inner solenoid, we need to determine the maximum change in magnetic flux through the inner solenoid.
The magnetic flux (Φ) through a solenoid is given by the formula Φ = μ₀*N*I*A, where μ₀ is the magnetic permeability of free space, N is the number of windings, I is the current, and A is the cross-sectional area of the solenoid.
First, let's find the magnetic flux through the inner solenoid:
The cross-sectional area of the inner solenoid (A_inner) can be calculated using the formula A_inner = π*(R_inner)^2, where R_inner is the radius of the inner solenoid.
Substituting the given values, we have A_inner = π*(0.02 m)^2 = 0.00126 m².
Now, we can calculate the flux through the inner solenoid:
Φ_inner = μ₀*N_inner*I_outer*A_inner,
where N_inner is the number of windings in the inner solenoid and I_outer is the current flowing through the outer solenoid.
Substituting the given values, we have Φ_inner = (4π*10^(-7) T*m/A)*(150)*(0.6 A)*(0.00126 m²).
After performing the calculations, we find that Φ_inner ≈ 0.000566 T*m².
The peak emf (ε_peak) in the inner solenoid is equal to the negative rate of change of the magnetic flux:
ε_peak = -dΦ_inner/dt.
Since the current in the outer solenoid is given by I(t) = I₀*sin(ωt), the rate of change of the current is found by taking the derivative of I(t) with respect to time:
dI(t)/dt = I₀*ω*cos(ωt).
Substituting the given values, we have dI(t)/dt = (0.6 A)*(100 s^(-1))*cos(100t).
Finally, we can calculate the peak emf in the inner solenoid:
ε_peak = -(dΦ_inner/dt) = -(0.000566 T*m²/s)*(0.6 A)*(100 s^(-1))*cos(100t).
So, the peak emf in the inner solenoid is given by the expression:
ε_peak = -0.034 T*m²/s*cos(100t) Volts.