A ball with a mass of 170 g which contains 4.20×108 excess electrons is dropped into a vertical shaft with a height of 110 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. Part 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. F F = nothing N Request Answer Part B Find the direction of the force that this magnetic field exerts on the ball just as it enters the field. 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

Since the ball starts from a height of h=110 m and has a vertical acceleration of g=9.8 m/s^2 (acceleration due to gravity), the final velocity can be found using the equation

[tex]v^2 = u^2 +2gh[/tex]

where u=0 is the initial velocity. Solving for v,

[tex]v=\sqrt{2(9.8 m/s^2)(110 m)}=46.4 m/s[/tex]

The ball contains

[tex]N=4.20\cdot 10^8[/tex] excess electrons, each of them carrying a charge of magnitude [tex]e=1.602\cdot 10^{-19}C[/tex], so the magnitude of the net charge of the ball is

[tex]Q=Ne=(4.20\cdot 10^8)(1.602\cdot 10^{-19}C)=6.72\cdot 10^{-11}C[/tex]

And given the magnetic field of strength B=0.250 T, we can now find the magnitude of the magnetic force acting on the ball:

[tex]F=qvB=(6.72\cdot 10^{-11}C)(46.4 m/s)(0.250 T)=7.8\cdot 10^{-10} N[/tex]

B) From north to south

The direction of the magnetic force can be found by using the right-hand rule:

- index finger: direction of motion of the ball -> downward

- middle finger: direction of magnetic field --> from east to west

- thumb: direction of the force --> from south to north

However, this is valid for a positive charge: in this problem, the ball is negatively charged (it is made of an excess of electrons), so the direction of the force is reversed: from north to south.

onyebuchinnaji onyebuchinnaji · Answer 2 · 2021-10-28T20:38:02+02:00

The magnitude of the force that the magnetic field exerts on the ball is 7.812 x 10⁻¹⁰ N.

The magnetic force will directed north to south (reverse direction due to charge of ball).

The given parameters;

mass of the ball, m = 170 g = 0.17 kg
excess electron, N = 4.2 x 10⁸ electrons
vertical height of the shaft, h = 110 m
magnetic field strength, B = 0.25 T

The speed of the charge is calculated by applying the principle of conservation of mechanical energy;

Potential energy at top = kinetic energy at bottom

mgh = ¹/₂mv²

2gh = v²

[tex]v = \sqrt{2gh} \\\\v= \sqrt{2\times 9.8 \times 110} \\\\v = 46.43 \ m/s[/tex]

The charge of the excess electron is calculated as follows;

Q = Ne

[tex]Q = (4.2 \times 10^8)\times (1.602 \times 10^{-19})\\\\Q = 6.73 \times 10^{-11} \ C[/tex]

The magnitude of the force that the magnetic field exerts on the ball is calculated as follows;

[tex]F = qvB\\\\F = 6.73 \times 10^{-11} \times 46.43 \times 0.25\\\\F = 7.812 \times 10^{-10} \ N[/tex]

The direction of the magnetic force will be perpendicular to speed of the charge and magnetic field.

The magnetic field is directed to east, the magnetic force will directed south to north. Since the ball is negatively charged (excess electron), the direction will be reversed. Thus, the force will be directed from north to south.

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