A ball with a mass of 120 g which contains 4.40×10 8 excess electrons is dropped

Question

A ball with a mass of 120 g which contains 4.40×108 excess electrons is dropped into a vertical shaft with a height of 120 m . At the bottom of the shaft, the ball suddenly enters a uniform horizontal magnetic field that has a magnitude of 0.260 T and direction from east to west.

a.) If air resistance is negligibly small, find the magnitude of the force that this magnetic field exerts on the ball just as it enters the field.

Use 1.602×1019 C for the magnitude of the charge on an electron.

b.) Find the direction of the force that this magnetic field exerts on the ball just as it enters the field.

1. From north to south

or

2. from south to north

Explanation / Answer

F = qvb sin theta

theta = 90 degree

sin90 =1

F = qvB

q = charge due to excess electron = N*q

q = 4.4 x 10^8 * 1.602 x 10^-19 = 7.04 x 10^-11 C

apply energy conservation

1/2*mv^2 = mgh

v = sqrt(2gh) = sqrt(2*9.8*120)

v = 48.49 m/s

B = 0.260 T

F = qvB = 7.04 x 10^-11 C * 48.49 m/s * 0.260 T =

F = 8.85 x 10^-10 N

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