Question

A calcium-40 ion has a positive charge that is double the charge of a proton, and a mass of 6.64 ✕ 10−26 kg. At a particular instant, it is moving with a speed of 4.95 ✕ 106 m/s through a magnetic field. At this instant, its velocity makes an angle of 56° with the direction of the magnetic field at the ion's location. The magnitude of the field is 0.160 T. (a) What is the magnitude of the magnetic force (in N) on the ion?

Answer 1

Answer:

F = 2.1 × $10^{-13}$ N

Explanation:

The force on a charge in a magnetic field is given by;

     F = qvBSin θ

Where: F is the force, q is the charge, v is the speed or velocity of the charge into the field, B is the magnetic field and θ is the angle with which the charge enters the field.

The charge on a proton = 1.6 × $10^{-19}$ C, so that q = 2 × 1.6× $10^{-19}$ = 3.2 × $10^{-19}$ C

Given that; q = 3.2 × $10^{-19}$ C, v = 4.95 × $10^{6}$ m/s, θ = 56°, B = 0.160 T.

Thus,

        F = 3.2 × $10^{-19}$ × 4.95 × $10^{6}$ × 0.160 × Sin 56°

           = 2.1 × $10^{-13}$ N

The magnitude of the magnetic force on the on is 2.1 × $10^{-13}$ N.