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Sagot :
Answer:
fem = 7.58 10⁻⁵ V
Explanation:
For this exercise we use Faraday's law
fem = [tex]- \frac{d \Phi _B}{dt}[/tex]
the magnetic flux is
Ф_B = B. A = B A cos θ
Tje bold are vectros. Suppose the case where the normal to the surface of the red blood cell is parallel to the field, therefore the angle is zero and the cos 0 = 1
The red blood cell area is
A =π r²
indicate that the diameter is r = 8.00 mm = 8.00 10⁻³ m
the magnetic field has a frequency of f=60 Hz, and B₀ = 1.00 10⁻³T, therefore we can write it
B = B₀ sin (wt) = B₀ sin( 2π f t)
we substitute
fem = - A dB / dt
fem = - A B₀ [tex]\frac{ d (sin ( 2\pi f t)}{dt}[/tex]
fem = - π r² Bo (2πf cos 2πft)
the maximum electromotive force occurs when the function is ±1
fem = 2 π² r² B₀ f
let's calculate
fem = 2π² (8.00 10⁻³)² 1.00 10⁻³ 60
fem = 7.58 10⁻⁵ V
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