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

Answer:

The diameter of the circular path decrease

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 $a_c=\frac{v^2}{r}$. Replacing and solving for B:

$qvB=m\frac{v^2}{r}\\B=\frac{mv}{qr}$

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.