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The body contains many small currents caused by the motion of ions in the organs and cells. Measurements of the magnetic field around the chest due to currents in the heart give values of about 10.0 μG . Although the actual currents are rather complicated, we can gain a rough understanding of their magnitude if we model them as a long, straight wire.If the surface of the chest is 5.40 cm from this current, how large is the current in the heart?

Respuesta :

Manetho

Answer:

270 μA

Explanation:

Use the magnetic field due to long, straight wire and solve for current I.

[tex]B= \frac{\mu_0I}{2\pi r}[/tex]

[tex]I= \frac{2\pi r B}{\mu_0}[/tex]

plug in the values

[tex]I= \frac{2\pi(5.40\times10^{-2})\times10\times10^{-6}\times10^{-4}}{4\pi\times10^{-7}}[/tex]

= 2.7×10^{-4)×10^6

=270 μA

The current that flows in the  heart is 270 μA

fichoh

Using the magnetic intensity relation, the current flowing in the heart based on the given parameters is [tex] I = 0.027 \times 10^{-3} \: Ampere[/tex]

Given the Parameters :

  • Radius, r = 5.40 cm = 0.054 m
  • Magnetic field intensity, B = 10μG = [tex] 10 \times 10^{-10} [/tex]

[tex] I = \frac{2πrB}{μ_{0}}[/tex]

Substituting the values into the formula to obtain the current, I :

I = \frac{2π(0.054)(10 \times 10^{-10}}{4π \times 10^{-7}}[/tex]

[tex] I = 0.0027 \times 10 \times 10^{-10+7} [/tex]

[tex] I = 0.027 \times 10^{-3} [/tex]

Therefore, the current flowing is [tex] 0.027 \times 10^{-3} \: Ampere[/tex]

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