Asingly charged positive ion has a mass of 3.46 × 10−26 kg. after being accelerated through a potential difference of 215 v the ion enters a magnetic field of 0.522 t in a direction perpendicular to the field. the charge on the ion is 1.602 × 10−19 c. find the radius of the ion’s path in the field. answer in units of cm.

1.8 cm

Explanation:

= mass of the singly charged positive ion = 3.46 x 10⁻²⁶ kg

= charge on the singly charged positive ion = 1.6 x 10⁻¹⁹ C

=Potential difference through which the ion is accelerated = 215 V

= Speed of the ion

Using conservation of energy

Kinetic energy gained by ion = Electric potential energy lost

= Radius of the path followed by ion

= Magnitude of magnetic field = 0.522 T

the magnetic force on the ion provides the necessary centripetal force, hence

f=ma newton's second law 7x8=56n