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 = 17.50 nT pointing north when the tornado was 9.10 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. A (conventional current) flowing ---Direction--- the tornado.

Answer 1

796.25 A

Step-by-step explanation:

B= (μI)/(2πr)

I= (B*2πr)/(μ)

μ= 4π*10^-7

I= ((17.50*10^-9)(2π)(9.10*10^3)) / (4π*10^-7)

= 796.25 A

A (conventional current) flowing "DOWN" the tornado.