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An incredible amount of electrical energy passes down the funnel of a large tornado every second. Measurements at a distance of 12.8 km from a large tornado showed an almost constant magnetic field of 2.35 × 10−8 T associated with the tornado. What was the average current going down the funnel? (µ0 = 4π × 10−7 T⋅m/A)

Respuesta :

Answer:

The average current going down the funnel is 1504 A.

Explanation:

Given that,

Magnetic field, [tex]B=2.35\times 10^{-8}\ T[/tex]

Measurements at a distance of 12.8 km from a large tornado, r = 12.8 km

We need to find the average current going down the funnel. We know that the magnetic field due to current carrying conductor is given by :

[tex]B=\dfrac{\mu_o I}{2\pi r}[/tex]

I is current

[tex]I=\dfrac{2\pi rB}{\mu_o}\\\\I=\dfrac{2\pi \times 12.8\times 10^3\times 2.35\times 10^{-8}}{4\pi \times 10^{-7}}\\\\I=1504\ A[/tex]

So, the average current going down the funnel is 1504 A. Hence, this is the required solution.

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