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 440 V. Determine the magnitude of the magnetic field, if the proton travels in a circular path with a radius of 17 cm. mT

Answer 1

Answer:

B = 17.8 mT

Explanation:

To find the velocity of the proton let's use conservation of energy

starting point. At the proton's exit point, from rest

         Em₀ = U = q V

final point. At the entry point of the magnetic field

         Em_f = K = ½ m v²

how energy is conserved

         Em₀ = Em_f

          qV = ½ m v²

          v = $\sqrt{2qV/m}$

let's calculate

           v = $\sqrt{ \frac{2 \ 1.6 \ 10^{-19} 440}{1.67 \ 10^{-27}} }$

           v = $\sqrt{843.1137 \ 10^8}$  

            v = 29.036 10⁴ m / s

Now let's use Newton's second law in the part where there is magnetic field

           F = ma

the force is magnetic

           F = q vxB

in this case the field is perpendicular to the velocity, therefore the magnitude of this formula,

           F = q vB

acceleration is centripetal

           a = v² / r

we substitute

           qvB = m v² / r

            B = $\frac{ mv}{ rq}$

let's calculate

            B = $\frac{ 1.67 \ 10^{-27} 29.036 \ 10^4}{ 0.17 \ 1.6 \ 10^{-19}}$

            B = 1.78 10⁻² T

let's reduce to mT

            B = 1.78 10⁻² T (1000mT / 1T)

            B = 17.8 mT