Respuesta :
Answer:
Explanation:
Flux
Φ = B A
= πR²B
-dΦ / dt = -πR²dB/ dt
EMF = - πR².00275d.cos ( ω t )/ dt
EMF = ( πR².00275/ω) sin ω t
EMFmax = ( πR².00275/ω)
EMFmax = ∫ E dR , E is elecreic field at distance R from the axis
= 2πR X E
( πR².00275/ω) = 2πR X Emax
Emax = R x .00275/2ω
= 1.85 x 10⁻² x .00275 / (2 x 349)
= .7288 x 10⁻⁷ N/C
= 7.288 x 10⁻⁸ N/C
Faraday's Law of Induction allows finding the result for the maximum electric field at a distance from the solenoid axis is:
E₀ = 4.685 10⁻³ V/s
Faraday's induction law says that the induced electromotive force is equal to minus the derived respect of time of the magnetic flux.
EMF = [tex]- \frac{d \Phi }{dt}[/tex]
The flow is
Ф = B . A
Where the bold indicate vectros, EMF is the induced electromotive force, Ф the magnetic flux, B the magnetic field and A the area.
Indicates that the magnetic field has the form
B = B₀ cos wt
With B₀ = 0.00275 T and w = 349 rad / s.
Solenoid area is constant, r= 4.25 cm.
A = pi r²
Let's do the derivative.
[tex]\frac{d \Phi }{dt } = A \frac{dB}{dt} \\\frac{d \Phi }{dt } = - A \ B_o \ w \ sin wt[/tex]
We substitute in the Faraday equation.
EMF = A B₀ w sint wt
They ask the maximum EMF is when the sine function is maximum.
sin wt = 1.
EMF₀ = A B₀ w (1)
Electric potential and electric field are related.
V = - ∫ E. ds
The electric field is constant, it can leave the integral, the radius is the distance R = 1.85 cm.
V = - E (2πR) (2)
We equate the expressions 1 and 2.
A B₀ w = E₀ 2π R
[tex]E_o = \frac{A \ B_o \ w }{2\pi \ R} \\E_o = \frac{r^2 B_o w }{2 R}[/tex]
Let's calculate.
E₀ = [tex]\frac{(4.25 \ 10^{-3})^2 \ 0.0275 349 }{2 \ 1.85 10^{-3} }[/tex]
E₀ = 4.685 10⁻³ V/m
In conclusion using Faraday's induction Law we can find the result for the maximum electric field at a distance from the solenoid axis is:
E₀ = 4.685 10⁻³ V/s
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