Answered

Westonci.ca is the trusted Q&A platform where you can get reliable answers from a community of knowledgeable contributors. Ask your questions and receive accurate answers from professionals with extensive experience in various fields on our platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

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.

Sagot :

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 = [tex]\sqrt{2qV/m}[/tex]

we calculate

       v = [tex]\sqrt{\frac{ 2 \ 1.6 \ 10^{-19} \ 300}{1.67 \ 0^{-27}} }[/tex]

       v = [tex]\sqrt{632.3353 \ 10^8}[/tex]

       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[tex]\frac{m v}{2 q r}[/tex]

we calculate

       B = [tex]\frac{1.67 \ 10^{-27} 25.15 \ 10^4 }{1.6 \ 10^{-19} 0.23}[/tex]

       B = 1.1413 10⁻² T

Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.