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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?

Respuesta :

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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

[tex]F= IlBsin\theta[/tex]

Where,

[tex]\theta =[/tex] 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

[tex]\frac{F}{l} = IBsin\theta[/tex]

Replacing with our values

[tex]\frac{F}{l} = (20000)(3*10^{-5})sin(90)[/tex]

[tex]\frac{F}{l} = 0.6N/m[/tex]

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

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