To monitor the breathing of a hospital patient, a thin belt isgirded around the patient's chest. The belt is a200-turn coil. When the patient inhales,

asked 2021-04-23
To monitor the breathing of a hospital patient, a thin belt isgirded around the patient's chest. The belt is a200-turn coil. When the patient inhales, the area encircled by the coil increases by \(\displaystyle{39.0}{c}{m}^{{2}}\). The magnitude of the Earth's magnetic field is 50.0uT and makes an angle of 28.0 degree with the plane of the coil. Assuming a patien takes 1.80s toin hale, find the magnitude of the average induced emf in the coilduring that time.
Do I use the equation \(\displaystyle{E}={N}\cdot{A}\cdot{B}{w}{\sin{{w}}}{t}\)?

Answers (1)

You were close. This is Faraday's Law of Induction.
The flux is \(\displaystyle\phi={N}{B}{A}{\cos{\theta}}\)
N is the number of turns
B is the strength of the magnetic field (should be in T)
A is the area of the loop (should be in \(\displaystyle{m}^{{2}}\))
The angle is between the magnetic field and the normal(perpendicular to the coil) so it would be cos(90.0 - 28.0).
The EMF = - rate of change in flux, or
The \(\displaystyle\triangle\phi={N}{B}{\left(\triangle{A}\right)}{\cos{\theta}}\) (since A is the only thingthat changes, and you don't need the original A, only the change in A)
If the above still confuses you...
Flux Linkage, or simply flux, refers to the magnetic field linespassing through the area of a loop:
The diagram on the left shows the loop allows the magnetic fieldvectors to pass through it (\(\displaystyle\theta\) ˜ 0 deg), but on the right,the magnetic field vectors cannot pass through it (\(\displaystyle\theta\) = 90deg).
The flux is \(\displaystyle\phi={N}{B}{A}{\cos{\theta}}\), so in the case on the cos (0deg) = 1, so \(\displaystyle\phi={N}{B}{A}\), but on the right cos (90 deg) = 0, so \(\displaystyle\phi={0}\).
The way that \(\displaystyle\theta\) is defined is it is perpendicular to the planeof the loops. Think about how the diagram on the right has a \(\displaystyle\theta\)= 90 deg. The area vector is pointing straight out at us in theright figure.
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