To monitor the breathing of a hospital patient, a thin belt is girded around the

Question

To monitor the breathing of a hospital patient, a thin belt is girded around the patient's chest as in the figure below. The belt is a 205-turn coil. When the patient inhales, the area encircled by the coil increases by 43.0 cm2. The magnitude of Earth's magnetic field is 50.0 T and makes an angle of 26.0° with the plane of the coil. Assuming a patient takes 1.90 s to inhale, find the magnitude of the average induced emf in the coil during that time. l = 9.66e-8 Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. Hv

Explanation / Answer

EMF - N*d(phi)/dt

phi = B.A = B*A*cos theta

|E| = N*B*cos theta*dA/dt

using given values:

N = 205

B = 50*10^-6 T

cos theta = 90 - 26 = 64 deg

dA/dt = 43810^-4/1.90 sec

So,

|E| = 205*50*10^-6*cos 64 deg*43*10^-4/1.90

|E| = 1.02*10^-5 V

Let me know if you have any doubt.

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