Question

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.

Answer 1

`F=9.09* 10^(-7)\ N`

Step-by-step 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 :

`mgh=(1)/(2)mv^2\\\\v=√(2gh) \\\\v=√(2* 9.8* 0.3) \\\\v=2.424\ m/s`

Now, the magnetic force is given by :

`F=qvB\\\\=0.25* 10^(-6)* 2.424* 1.5\\\\=9.09* 10^(-7)\ N`

So, the magnetic force is
`9.09* 10^(-7)\ N`. Since, the bob is at the lowest point, the direction of the magnetic force at the lowest point is upward.