Question

A ball with a mass of 170 g which contains 3.80×108 excess electrons is dropped into a vertical shaft with a height of 145 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 1.602×10−19 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.

a-from north to south

b-from south to north

Answer 1

Answer:

A. F=6.65*10^{-10}N

B. south - north

Explanation:

A) We use the Lorentz force

F = qv X B

|F| = qvB

to calculate the magnitude of the force we need the speed of the of the ball.

$v_{f}^{2}=v_{0}^{2}+2gy\\v_{f}=\sqrt{0+2(9.8\frac{m}{s^{2}})(145m)}=53.31\frac{m}{s}$

and by replacing in the formula for the magnitude of the force we have (taking into account the excess of electrons)

$F=(3.8*10^{8})(1.602*10^{-19}C)(53.31\frac{m}{s})(0.205T)=6.65*10^{-10}N$

B)

b.  south - north (by the rigth hand rule)

I hope this is usefull for you

regards