Item 10 Part A A ball with a mass of 140 g which contains 3.70x108 excess electr

Question

Item 10 Part A A ball with a mass of 140 g which contains 3.70x108 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 0.250 T and direction from east to west. 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 1.602x10-19 C for the magnitude of the charge on an electron. Submit My Answers Give Up Part B Find the direction of the force that this magnetic field exerts on the ball just as it enters the field from north to south from south to north Submit My Answers Give Up Provide Feedback Continue

Explanation / Answer

charge on the ball = q = -n*e

velocity of the ball as it enters the magnetic field v = -sqrt(2*g*h) k

v = -sqrt(2*9.8*120) k = -48.5 k

magnetic forceFb = q*(v x B )

v = -48.5 k

B = -0.25 i

Fb = -3.7*10^8*1.6*10^-19*( -48.5 k x -0.25 i )

Fb = -3.7*10^8*1.602*10^-19*48.5*0.25 j

Fb = -7.2*10^-10 j N

part A

direction = 7.2*10^-10 N

part B

from north to south

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