A ball with a mass of 170 g which contains 3.90×10 8 excess electrons is dropped

Question

A ball with a mass of 170 g which contains 3.90×108 excess electrons is dropped into a vertical shaft with a height of 150 m . At the bottom of the shaft, the ball suddenly enters a uniform horizontal magnetic field that has a magnitude of 0.250 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.

Explanation / Answer

q = 3.9*10^8*e = 3.9*10^8*1.6*10^-19 = 6.24*10^-11 C

using energy conservation

speed at the bottom of the shaft

0.5*mv^2 = mgy

v = sqrt(2gy)

v = sqrt(2*9.81*150) = 54.25 m/sec

F = qvB*sin A

F = 6.24*10^-11*54.25*0.250*sin 90 = 8.463*10^-10 N

Direction

v is down ward

B is west

vxB = north direction

since q is negative

So F will be in south

Ans. From north to south

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