A140-g ball containing 4.10 * 108 excess electrons is dropped into a 110-m vertical shaft. at the bottom of the shaft, the ball suddenly enters a uniform horizontal magnetic field that has magnitude 0.300 t and direction from east to west. if air resistance is negligibly small, find the magnitude and direction of the force that this magnetic field exerts on the ball just as it enters the field.

Magnetic force, from north to south.

Explanation:

It is given that,

Mass of the ball, m = 140 g = 0.14 kg

Number of electrons,

Height, h = 110 m

Charge, q = ne

Magnetic field, B = 0.3 T

The potential energy at height of 110 meters is balanced by the kinetic energy as per the law of conservation of energy as :

Magnetic force is given by :

F = q v B

As the direction of magnetic field is from east to west, so from right hand thumb rule the direction of magnetic force is from north to south. Hence, this is the required solution.

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .

i think the answer is a, hope this is right, that would be my best guess.