To monitor the breathing of a hospital patient, a thin belt is girded around the patient's chest . The belt is a 215-turn coil. When the patient inhales, the area encircled by the coil increases by 41.0 cm2. The magnitude of Earth's magnetic field is 50.0 µT and makes an angle of 30.0° with the plane of the coil. Assuming a patient takes 2.10 s to inhale, find the magnitude of the average induced emf in the coil during that time.

|| = µV

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13 years ago

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Flux change/Time = N*B*A*sin30/2.1s
= N*(50*10^-6T)*(41*10^-4m^2)*(0.5)/2.1s
= (215)*1.95*10^-8 V
= 4.2 µV

answered by drwls

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Question

To monitor the breathing of a hospital patient, a thin belt is girded around the patient's chest . The belt is a 215-turn coil. When the patient inhales, the area encircled by the coil increases by 41.0 cm2. The magnitude of Earth's magnetic field is 50.0 µT and makes an angle of 30.0° with the plane of the coil. Assuming a patient takes 2.10 s to inhale, find the magnitude of the average induced emf in the coil during that time.
|| =   µV

drwls · Answer

Flux change/Time = N*B*A*sin30/2.1s = N*(50*10^-6T)*(41*10^-4m^2)*(0.5)/2.1s = (215)*1.95*10^-8 V = 4.2 µV