Respuesta :
- (a) Maximum emf: 90 V (2 sig. fig.)
- (b) Emf at π/32 s: 85 V.
- (c) t = 0.125 s.
Explanation
(a)
The maximum emf in the coil depends on
- the maximum flux linkage through the coil, and
- the angular velocity of the coil.
Maximum flux linkage in the coil:
[tex]\phi_\text{max} = B\cdot A\cdot N = 0.59\;\text{T}\times(0.08 \times 0.30)\;\text{m}^{2} \times 505 = 7.2\;\text{Wb}[/tex].
Frequency of the rotation:
[tex]f = 120\;\text{rev}\cdot\text{min}^{-1} = 2 \;\text{rev}\cdot\text{s}^{-1}[/tex].
Angular velocity of the coil:
[tex]\omega = 2\;\pi\;\text{rev}^{-1}\times 2\;\text{rev}\cdot\text{s}^{-1} = 4 \pi \;\text{s}^{-1}[/tex].
Maximum emf in the coil:
[tex]\epsilon_\text{max} = \omega\cdot\phi_\text{max} = 4\;\pi \times 7.2\;\text{Wb} = 90\;\text{V}[/tex].
(b)
Emf varies over time. The trend of change in emf over time resembles the shape of either a sine wave or a cosine wave since the coil rotates at a constant angular speed. The question states that emf is "zero at t = 0." As a result, a sine wave will be the most appropriate here since [tex]\sin{0} = 0[/tex].
[tex]\displaystyle \epsilon(t) = \epsilon_\text{max}\cdot \sin{(\omega\cdot t)}[/tex].
Make sure that your calculator is in the radian mode.
[tex]\displaystyle \epsilon\left(\frac{\pi}{32}\right) = 90\;\text{V}\times \sin\left(4\;\pi\times \frac{\pi}{32}\right) = 85\;\text{V}[/tex].
(c)
Consider the shape of a sine wave. The value of [tex]\displaystyle \sin\left(\omega \cdot t\right)[/tex] varies between -1 and 1 as the value of [tex]t[/tex] changes. The value of [tex]\epsilon[/tex] at time [tex]t[/tex] depends on the value of [tex]\sin(\omega \cdot t)[/tex].
[tex]\sin(\omega \cdot t)[/tex] reaches its first maximum for [tex]t\ge 0[/tex] when what's inside the sine function is equal to [tex]\pi/2[/tex].
In other words, the first maximum emf occurs when
[tex]\omega \cdot t = \dfrac{\pi}{2}[/tex],
where
[tex]\sin{\omega \cdot t} = 1[/tex],
and
[tex]\epsilon = \epsilon_\text{max}[/tex].
[tex]\displaystyle t = \frac{\pi}{2}/\omega = \frac{1}{8} = 0.125\;\text{s}[/tex].
