Question

A proton moves in a circular path perpendicular to a constant magnetic field. If the field strength of the magnet is increased, does the diameter of the circular path increase, decrease, or remain the same?

Answer 1

The diameter of the circular path decrease

Step-by-step explanation:

The magnitude of the magnetic force exerted on the proton is given by:

`F=qvB`

According to Newton's second law, this force is equal to the centripetal force. Therefore, we have:

`F=F_c\\qvB=ma_c`

Recall that
`a_c=(v^2)/(r)`. Replacing and solving for B:

`qvB=m(v^2)/(r)\\B=(mv)/(qr)`

Thus, the field stregth is inversely proportional to the radius of the path and since the radius is directly proportional to the diameter, if the field strength increases, the diameter decreases.