Question

In 1962 measurements of the magnetic field of a large tornado were made at the Geophysical Observatory in Tulsa, Oklahoma. If the magnitude of the tornado's field was B = 1.55 ✕ 10−8 T pointing north when the tornado was 8.55 km east of the observatory, what current was carried up or down the funnel of the tornado? Model the vortex as a long, straight wire carrying a current.

Answer 1

Answer:

Explanation:

we have:

$B = 1.55 \times 10^{-8} T\\r = 8.55 km= 8.55 x 10^3m$

We know:

$B= \frac{\mu_0I}2\pi r}\\\\\therefore I =\frac{2\pi rB}{\mu_0}... (\mu_0 = 4\pi \times 10^{-7} T-m/A)\\\\= \frac{[2\pi ( 8.55 \times 10^3m)(1.55 \times 10^{-8} T)]}{4\pi \times 10^{-7} T-m/A}\\\\ =662.625 A$