Answer:
Explanation:
First of all, we have to take into account that the emf induced in a solenoid is given by
emf = -n*ΔФ/Δt
where n is the number of turns of wire per cm and ΔФ is the change in the magnetic flux. Hence, we have to calculate ΔФ by taking into account the magnetic field produced by the solenoid
[tex]B=\mu n I = (4\pi *10^{-7}\frac{Tm}{A})(\frac{87.2}{10^{-2}m})(3.2A)=0.03T[/tex]
This is the magnetic field that crosses the secondary winding. Now, we have to compute the flux
[tex]\Phi = BA=(0.03T)(1.94*10^{-4}m^{2})=6.8*10^{-6}Tm^{2}[/tex]
and for the calculation of the change in time we have to take t from the expression for i(t)
[tex]i(t)=(0.160A/s^{2})t^{2}=3.2A\\t=\sqrt{\frac{3.2A}{0.160A/s^{2}}}=4.47s[/tex]
Hence, the emf in the secondary winding (with n=5) is:
[tex]emf=-(5)\frac{6.8*10^{-6}Tm^{2}-0}{4.47s-0}=7.6*10^{-6}V[/tex]
I hope this is usefull for you
regards
Given Information:
current = I = 3.2 A
Area of solenoid = A = 1.94 cm² = 0.94x10⁻³ m²
Turns per cm at primary side = n = 87.2 turns/cm = 8720 turns/m
current at time 0 = i(t) = (0.160 A/s²) t²
Number of turns secondary side = N = 5 turns
Required Information:
Induced emf at secondary side = ξ = ?
Answer:
Induced emf at secondary side ξ = -7.605x10⁻⁶ V
Explanation:
From the Faraday's law the induced EMF ξ is given by
ξ = -NΔΦ/Δt
t can be found using the given i(t)
3.2 = 0.160 t²
t² = 3.2/0.160
t = √3.2/0.160
t = 4.472 seconds
change in flux ΔΦ is given by
ΔΦ = BA
ΔΦ = μ₀nIA
ΔΦ = 4πx10⁻⁷*8720*3.2*0.194x10⁻³
ΔΦ = 6.802x10⁻⁶ Tm²
So the induced emf in the secondary winding is
ξ = -NΔΦ/Δt
ξ = -5*6.802x10⁻⁶/4.472
ξ = -7.605x10⁻⁶ V
The negative sign indicates that the induced emf opposes the change that produced it in the first place.
