Question

A proton enters a region of constant magnetic field, perpendicular to the field and after being accelerated from rest by an electric field through an electrical potential difference of 330 V. Determine the magnitude of the magnetic field, if the proton travels in a circular path with a radius of 23 cm.

Answer 1

Answer:

 B = 1.1413 10⁻² T

Explanation:

We use energy concepts to calculate the proton velocity

starting point. When entering the electric field

        Em₀ = U = q V

final point. Right out of the electric field

        em_f = K = ½ m v²

energy is conserved

       Em₀ = Em_f

       q V = ½ m v²

       v = $\sqrt{2qV/m}$

we calculate

       v = $\sqrt{\frac{ 2 \ 1.6 \ 10^{-19} \ 300}{1.67 \ 0^{-27}} }$

       v = $\sqrt{632.3353 \ 10^8}$

       v = 25.15 10⁴ m / s

now enters the region with magnetic field, so it is subjected to a magnetic force

        F = m a

the force is

       F = q v x B

as the velocity is perpendicular to the magnetic field

       F = q v B

acceleration is centripetal

       a = v² / r

we substitute

       qvB =1/2  m v² / r

       B =  v$\frac{m v}{2 q r}$

we calculate

       B = $\frac{1.67 \ 10^{-27} 25.15 \ 10^4 }{1.6 \ 10^{-19} 0.23}$

       B = 1.1413 10⁻² T