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Sagot :
Given:
• Mass, m = 6.64 x 10⁻²⁷ kg
,• Velocity = 42.3 km/s
,• Magnetic field = 1.60 T
Let's find the diameter of the path followed by the particle.
To find the path, apply the formula to first find the radius:
[tex]R=\frac{mv}{QB}[/tex]Where:
m = 6.64 x 10⁻²⁷ kg
v is the velocity in m/s = 42.3 x 1000 = 42300 m/s
Q = 1.6 x 10⁻¹⁹ x 2
B = 1.60 T
Plug in values and solve for the radius, r:
[tex]\begin{gathered} R=\frac{6.64*10^{-27}*42300}{(3.2*10^{-19})*1.60} \\ \\ R=5.48578*10^{-4}\text{ m} \end{gathered}[/tex]Therefore, the diameter will be:
[tex]\text{ diameter = radius x 2 = \lparen5.48578*10}^{-4})*2=0.00101\text{ m}\approx1.01\text{ mm}[/tex]Therefore, the diameter of the path followed by the particle is 1.01 mm.
ANSWER:
1.01 mm
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