Question

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)

Answer 1

The average current going down the funnel is 1504 A.

Step-by-step explanation:

Given that,

Magnetic field,
`B=2.35* 10^(-8)\ T`

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 :

`B=(\mu_o I)/(2\pi r)`

I is current

`I=(2\pi rB)/(\mu_o)\\\\I=(2\pi * 12.8* 10^3* 2.35* 10^(-8))/(4\pi * 10^(-7))\\\\I=1504\ A`

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