Question

A positively charged particle of mass 6.20 10-8 kg is traveling due east with a speed of 75 m/s and enters a 0.47-T uniform magnetic field. The particle moves through one-quarter of a circle in a time of 6.40 10-3 s, at which time it leaves the field heading due south. All during the motion the particle moves perpendicular to the magnetic field.(a) what is the magnitude of the magnetic force on the particle?(b) what is the charge on the particle?

Answer 1

Answer:

$F=1.143*10^{-3}N$ and $q=3.24*10^{-5}c$

Explanation:

a) We need first to identify the Period from the one-quarte 'time' given. That is

$t=6.4*10^{-3}s$

Then the period is,

$T=4t=4(6.4*10^{-3})=25.6*10^{-3}$

From this Period we can calculate the radio,

$v=\frac{2\pi r}{T}$

Rearrange for r,

$r=\frac{vT}{2\pi}\\r=\frac{(75m/s)(25.6*10^{-3}s)}{2\pi}\\r=0.305m$

We know can calculate the Magnetic force,

$F=\frac{mv^2}{r}=\frac{(6.2*10^{-8})(75)^2}{0.305}\\F=1.143*10^{-3}N$

b) To find the charge we only use the Force in terms of the electric field, that is

$F=Bvq$

Rearrange to q,

$q=\frac{F}{Bv}\\q=\frac{1.143*10^{-3}}{(0.47)(75)}\\q=3.24*10^{-5}c$