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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
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