Question

A positively charged particle of mass 7.2 x 10-8 kg is traveling due east with a speed of 88 m/s and enters a 0.6-T uniform magnetic field. The particle moves through one-quarter of a circle in a time of 2.2 x 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. Determine the magnitude of its charge in mC. (1 mC

Answer 1

Answer:

q = 8.57 10⁻⁵ mC

Explanation:

For this exercise let's use Newton's second law

         F = ma

where force is magnetic force

        F = q v x B

the bold are vectors, if we write the module of this expression we have

         F = qv B sin θ

as the particle moves perpendicular to the field, the angle is θ= 90º

        F = q vB

the acceleration of the particle is centripetal

        a = v² / r

we substitute

        qvB = m v² / r

         qBr = m v

          q =$\frac{m\ v}{B\ r}$

The exercise indicates the time it takes in the route that is carried out with constant speed, therefore we can use

          v = d / t

the distance is ¼ of the circle,

          d = $\frac{1}{4} \ 2\pi r$

           d =$\frac{\pi }{2r}$

we substitute

           v =  $\frac{\pi r}{2t}$

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

           

let's calculate

           r =$\frac{2 \ 2.2 \ 10^{-3} \ 88}{\pi }$ 2 2.2 10-3 88 /πpi

           r = 123.25 m

         

let's substitute the values

           q = $\frac{ 7.2 \ 10^{-8} \ 88}{ 0.6 \ 123.25}$7.2 10-8 88 / 0.6 123.25

            q = 8.57 10⁻⁸ C

Let's reduce to mC

           q = 8.57 10⁻⁸ C (10³ mC / 1C)

           q = 8.57 10⁻⁵ mC