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

Step-by-step explanation:

we have:

`B = 1.55 *  10^(-8) T\\r = 8.55 km= 8.55 x 10^3m`

We know:

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