Respuesta :
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 = [tex]\sqrt{2qV/m}[/tex]
let's calculate
v = [tex]\sqrt{ \frac{2 \ 1.6 \ 10^{-19} 440}{1.67 \ 10^{-27}} }[/tex]
v = [tex]\sqrt{843.1137 \ 10^8}[/tex]
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 = [tex]\frac{ mv}{ rq}[/tex]
let's calculate
B = [tex]\frac{ 1.67 \ 10^{-27} 29.036 \ 10^4}{ 0.17 \ 1.6 \ 10^{-19}}[/tex]
B = 1.78 10⁻² T
let's reduce to mT
B = 1.78 10⁻² T (1000mT / 1T)
B = 17.8 mT
