Question

A lightening bolt may carry a current of 11800 A for a short period of time. The permeability of free space is 1.25664 × 10−6 T · m/A. What is the resulting magnetic field 47 m from the bolt? Suppose that the bolt extends far above and below the point of observation.

Answer 1

Answer:

The resulting magnetic field is 5.021 x 10⁻⁵ T

Explanation:

Given;

current in the lightening bolt, I = 11800 A

distance from the bolt, r = 47 m

permeability of free space, μ₀ = 1.25664 × 10⁻⁶ T· m/A

Assume lightening bolt as long straight conductor, then the resulting magnetic field will be calculated as follows;

$B = \frac{\mu_oI}{2 \pi r}$

where;

B is the resulting magnetic field

I is the current in the bolt

r is the distance from the bolt

Substitute the given values and calculate B

$B = \frac{\mu_oI}{2 \pi r} \\\\B = \frac{1.25664 *10^{-6}*11800}{2 \pi (47)} \\\\B = 5.021 *10^{-5} \ T$

Thus, the resulting magnetic field is 5.021 x 10⁻⁵ T