Answer :
Answer:
q = 8.57 10⁻⁵ mC
Explanation:
For this exercise let's use Newton's second law
F = ma
where force is magnetic force
F = q v x B
the bold are vectors, if we write the module of this expression we have
F = qv B sin θ
as the particle moves perpendicular to the field, the angle is θ= 90º
F = q vB
the acceleration of the particle is centripetal
a = v² / r
we substitute
qvB = m v² / r
qBr = m v
q =[tex]\frac{m\ v}{B\ r}[/tex]
The exercise indicates the time it takes in the route that is carried out with constant speed, therefore we can use
v = d / t
the distance is ¼ of the circle,
d = [tex]\frac{1}{4} \ 2\pi r[/tex]
d =[tex]\frac{\pi }{2r}[/tex]
we substitute
v = [tex]\frac{\pi r}{2t}[/tex]
r = [tex]\frac{2 \ t \ v}{\pi }[/tex]
let's calculate
r =[tex]\frac{2 \ 2.2 \ 10^{-3} \ 88}{\pi }[/tex] 2 2.2 10-3 88 /πpi
r = 123.25 m
let's substitute the values
q = [tex]\frac{ 7.2 \ 10^{-8} \ 88}{ 0.6 \ 123.25}[/tex]7.2 10-8 88 / 0.6 123.25
q = 8.57 10⁻⁸ C
Let's reduce to mC
q = 8.57 10⁻⁸ C (10³ mC / 1C)
q = 8.57 10⁻⁵ mC