Question

A ball with a mass of 160 g which contains 4.40×108 excess electrons is dropped into a vertical shaft with a height of 115 m . At the bottom of the shaft, the ball suddenly enters a uniform horizontal magnetic field that has a magnitude of 0.200 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 1.602×10−19 C for the magnitude of the charge on an electron.

Answer 1

Answer:

$F= 6.69*10^{-10}N$

Explanation:

To find the force we proceed by defining the variables we have,

$m= 160g\\n= 4.40*10^8\\h=115m\\B=0.2T\\e= 1.602*10^{-19}c$

The charge on one of the balls is defined under the equation,

$q=ne$

$q= 4.40*10^8(1.602*10^{-19}c)$

$q= 7.0488*10^{-11}$

Due to the height we need to calculate the potential energy at the height of 115m,

$PE=mgh$

The kinetic energy would be given by

$KE= \frac{1}{2}mv^2$

From the law of conservation we equate the two equations

$\frac{1}{2}mv^2=mgh$

$v=\sqrt{2gh}$

$v=\sqrt{2(9.8)(115)}$

$v= 47.47m/s$

In this way we now calculate the strength of the particle

$F=qVB$

$F= (7.0488*10^{-11})(47.47)(0.2)$

$F= 6.69*10^{-10}N$