What is the magnitude and direction of the magnetic force on the bob at the lowest point in its path, if it has a positive 0.250 μC charge and is released from a height of 30.0 cm above its lowest point? The magnetic field strength is 1.50 T.

Explanation:

Given that,

Charge, q = 0.250 μC

It is released from a height of 30 cm or 0.03 m

The magnetic field strength is 1.50 T.

First we find the velocity using the conservation of energy as follows :

Now, the magnetic force is given by :

So, the magnetic force is . Since, the bob is at the lowest point, the direction of the magnetic force at the lowest point is upward.

found in the nucleus

has mass of one amu

probably not the "magnets"in the magnet, so to speak, are thought to be electrons orbiting in shells. vurrent loops two sides to the loop