A 44-turn rectangular coil with length ℓ = 17.0 cm and width w = 8.10 cm is in a region with its axis initially aligned to a hor
1 answer:
Answer:
The maximum induced emf in the rotating coil = 29.66V
The induced emf in the rotating coil when (t = 1.00 s) = 26.66V
The maximum rate of change of the magnetic flux through the rotating coil = 0.674Wb/s
Explanation:
Lets state the parameters we are being given right from the question:
Number of rectangular coil, (N) = 44
Length of Coil, l =17cm in meters we have; (l) = 17 × 10⁻² m
Width of Coil, w =8.10cm in meters we have; (w) = 8.10 × 10⁻² m
Magnitude of Uniform Magnetic Field (B) = 767mT= 765 × 10⁻³ T
Angular Speed of Coil, (ω) = 64 rad/s
(a)
To calculate the induced emf in the rotating cell,we can use the formula:
emf = NBAωsin(ωt)
For maximum induced emf, the value of sin(ωt) will be 1
= NBAω ; if (A = l × w) , we have:
= NB(l × w)ω
subsitituting the parameters into the above equation; we have:
= 44 × 765 × 10⁻³ ( 17 × 10⁻² × 8.10 × 10⁻² ) × 64
= 29.66V
(b)
At t = 1s, the induced emf is calculated as:
emf = NBAωsin(ωt)
substituting the parameters into the equation, we have:
emf = 44 × 765 × 10⁻³ ( 17 × 10⁻² × 8.10 × 10⁻² ) × 64 × sin (64 × 1)
=26.66V
(c)
To calculate the maximum rate of change of the magnetic flux through the rotating coil; we need to reflect on the equation for the maximum induced emf in terms of magnetic flux.
i.e =
since = 29.66 and N = 44; we have:
29.66 =
=
= 0.674 Wb/s
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Answer:
a) I= I₀ (cos²θ - cos⁴θ) b) 75.5º
Explanation:
a) For this exercise we must use Malus's law
I = I₀ cos² θ
where tea is the angle between the two polarizers.
We apply this expression to our case
* Polarizer 1 suppose that it is vertical and polarizer 2 (intermediate) is at an angle θ with respect to the vertical
I₁ = I₀ cos² θ
* We analyze for the polarity 2 and the last polarizer 3 which indicate that it must be at 90º from the first one, therefore it must be horizontal.
The angle of polarizers 2 and 3 is θ' measured from the horizontal, if we measure with respect to the vertical
θ₂ = 90- θ’ = θ
fiate that in the exercise we must take a reference system and measure everything with respect to this system.
I = I₁ cos² θ'
we substitute
I = (I₀ cos² tea) cos² (θ - 90)
cos (θ -90) = cos θ cos 90 + sin θ sin 90 = sin θ
I = Io cos² θ sin² θ
1= cos²θ+ sin²θ
sin²θ = 1 - cos²θ
I= I₀ (cos²θ - cos⁴θ)
b) to find when the intensity is maximum,
we can use that we have an extreme point when the drift is zero
= 0
\frac{dI}{d \theta}= Io (2 cos θ - 4 cos³θ) = 0
whereby
cos θ - 2 cos³ θ = 0
cos θ ( 1 - 2 cos² θ) = 0
The zeros of this function are in
θ = 90º
1-2cos²θ =0 cos θ = 0.25 θ = 75.5º
Let's analyze this two results for the angle of 90º the intnesidd is zero with respect to the first polarizer, so it is not an acceptable solution.
Consequently, the angle that allows the maximum intensity to pass is 75.5º