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xxMikexx [17]
4 years ago
14

Scientific work is currently underway to determine whether weak oscillating magnetic fields can affect human health. For example

, one study found that drivers of trains had a higher incidence of blood cancer than other railway workers, possibly due to long exposure to mechanical devices in the train engine cab. Consider a magnetic field of magnitude 0.00100 T, oscillating sinusoidally at 61.5 Hz. If the diameter of a red blood cell is 7.20 µm, determine the maximum emf that can be generated around the perimeter of a cell in this field.
Physics
1 answer:
nlexa [21]4 years ago
6 0

Answer:

The maximum emf that can be generated around the perimeter of a cell in this field is 1.5732*10^{-11}V

Explanation:

To solve this problem it is necessary to apply the concepts on maximum electromotive force.

For definition we know that

\epsilon_{max} = NBA\omega

Where,

N= Number of turns of the coil

B = Magnetic field

\omega = Angular velocity

A = Cross-sectional Area

Angular velocity according kinematics equations is:

\omega = 2\pi f

\omega = 2\pi*61.5

\omega =123\pi rad/s

Replacing at the equation our values given we have that

\epsilon_{max} = NBA\omega

\epsilon_{max} = NB(\pi (\frac{d}{2})^2)\omega

\epsilon_{max} = (1)(1*10^{-3})(\pi (\frac{7.2*10^{-6}}{2})^2)(123\pi)

\epsilon_{max} = 1.5732*10^{-11}V

Therefore the maximum emf that can be generated around the perimeter of a cell in this field is 1.5732*10^{-11}V

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The  distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.

<h3>Distance from the center of the meter rule</h3>

The distance from the centre of the rule at which a 2N weight must be suspend from A is calculated as follows;

-----------------------------------------------------------------

  20 A  (30 - x)↓      x         ↓      20 cm  B 30 cm

                       2N              0.9N

Let the center of the meter rule = 50 cm

take moment about the center;

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(30 - x)(2 + 0.9x) = 18

60 + 27x - 2x - 0.9x² = 18

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0.9x² - 25x - 42 = 0

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Thus, the  distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.

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<u>Answer</u>:

(B) A pot being heated by an electric burner

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