In a model generator, a 310-turn rectangular coil 0.075 m by 0.18 m rotates with an angular frequency of 12.2 rad/s in a uniform
1 answer:
This question involves the concepts of the induced emf, magnetic field, and angular frequency.
The maximum emf induced in the coil is "32.68 volts".
The following formula can be used to find out the induced emf in the coil.
But, for maximum induced emf sin function should have a value of 1. Therefore,
where,
E = induced emf = ?
N = No. of turns in the coil = 310 turns
ω = angular frequency = 12.2 rad/s
A = Area of coil = (0.075 m)(0.18 m) = 0.0135 m²
B = magnetic field = 0.64 T
Therefore,
<u>E = 32.68 volts</u>
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Answer
Given,
refractive index of film, n = 1.6
refractive index of air, n' = 1
angle of incidence, i = 35°
angle of refraction, r = ?
Using Snell's law
n' sin i = n sin r
1 x sin 35° = 1.6 x sin r
r = 21°
Angle of refraction is equal to 21°.
Now,
distance at which refractive angle comes out
d = 2.5 mm
α be the angle with horizontal surface and incident ray.
α = 90°-21° = 69°
t be the thickness of the film.
So,
t = 2.26 mm
Hence, the thickness of the film is equal to 2.26 mm.