Question

What is the force per meter on a lightning bolt at the equator that carries 20,000 A perpendicular to Earth’s 3.0e−5 T field? (b) What is the direction of the force if the current is straight up and Earth’s field direction is due north, parallel to the ground?

Answer 1

To solve this problem it is necessary to apply the concepts related to Magnetic Force.

The magnetic force as magnitude of a vector is described by

$F= IlBsin\theta$

Where,

$\theta =$ Angle between the wire and the magnetic field

L = Length

B = Magnetic Field

I = Current

Since the lightning bolt is perpendicular to the earth then the angle is 90°, moreover we need find the force per meter, then

$\frac{F}{l} = IBsin\theta$

Replacing with our values

$\frac{F}{l} = (20000)(3*10^{-5})sin(90)$

$\frac{F}{l} = 0.6N/m$

Therefore the force per meter on lightning volt is 0.6N/m