Question

In a long, straight, vertical lightning stroke, electrons move downward and positive ions move upward and constitute a current of magnitude 20.0kA . At a location 50.0m east of the middle of the stroke, a free electron drifts through the air toward the west with a speed of 300 m/s . (e) If it does not collide with any obstacles, how many revolutions will the electron complete during the 60.0-μs duration of the lightning stroke?

Answer 1

The number of revolutions the electron completes in 60.0-μs of the strike is 134.

A magnetic field, a vector field that describes the magnetic influence on moving electric charges, electric currents, and magnetic materials. When a charge moves through a magnetic field, a force that is perpendicular to both its own velocity and the magnetic field operates on it.

Electrons go downward and positive ions move upward in a long, straight, vertical lightning stroke, creating a current of magnitude I = 20.0 kA.

A free electron travels through the air at a speed of v = 300 m/s at a place r = 50.0 m east of the stroke's center.

Let the magnetic field be B, and F be the magnetic force.

Counterclockwise horizontal arcs of field lines are produced by the upward lightning current.

We have, B = 8 × 10⁻⁵ T and;

The mass of an electron is, m = 9.11 × 10⁻³¹ kg

The time interval is Δt = 60 μs = 60 × 10⁻⁶

The angular frequency is given as:

ω = qB /m = 2πN / Δt

Where the number of revolutions is N.

So,

N = qBΔt /2πm

N = (l.60 × l0⁻¹⁹)(8 × l0⁻⁵)(60 × 10⁻⁶) / 2π(9.11 × 10⁻³¹ kg)

N = 134 revolutions

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