Assuming a patient takes 1.80 s to inhale, find the average induced emf in the coil during this time interval.

+3 votes

To monitor the breathing of a hospital patient, a thin belt is girded around the patient's chest. The belt is a 200-turn coil. When the patient inhales, the area encircled by the coil increases by 47.0 cm2. The magnitude of the Earth's magnetic field is 50.0 μT and makes an angle of 28.0° with the plane of the coil. Assuming a patient takes 1.80 s to inhale, find the average induced emf in the coil during this time interval. 


 µV

asked Feb 28, 2012 in Physics by awesome ~Expert~ (1,479 points)
    

1 Answer

+1 vote

Emf =  - dΦB/dt = - d/dt (B*A*N)*sin(theta) (*where B=magnetic field, A=area, N=number of turns)

B = 50μT  = 50*10^-6 T

A = 47 cm^2 = 0.0047 m^2

N = 200 turns

d/dt = 1/1.8 s

theta = 28

plugging back into the formula we get:

Emf =- (1/1.8 s)*(50*10^-6 T)(0.0047 m^2)(200) *sin(28)

Emf = -1.23*10^-5 V = 12.3 µV

answered Mar 8, 2012 by pokemonmaster ~Expert~ (3,856 points)



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