Answer:
The radius described the proton moving is
r=3.91mm
Explanation:
The magnetic field that moving the proton with the velocity
v=15 km/s
To find the radius combinate the magnetic field with the mass of the proton and the charge of the proton
r = mv=eB
r = mv/eB
[tex]r=\frac{(1.67*10^{-27})*(15\frac{km}{s})}{(1.6*10^{-19})(4*10^-2kg)}[/tex]
r = 3.9mm
The radius of the path will be:
0.0000039 km = 0.0039 m
For a charged particle of charge q, moving with velocity v on a perpendicular magnetic field of intensity B, the radius of the motion will be:
[tex]r = \frac{1}{B}*\frac{mv}{q} }[/tex]
Where m is the mass of the particle.
For a proton we have:
And we know that:
B = 400 G.
v = 15km/s
Then the radius is given by:
[tex]r = \frac{1}{400G}*\frac{*(1.673*10^{(-27)} kg)(15km/s}{1.602*10^{(-19)} C} }\\\\r = \frac{1}{400G}*1.567*10^{-7} kg*km/s*C}[/tex]
1 G = Mx/cm^2
We now can rewrite:
400G = 400 mx/cm^2 = 4*10^(-6) Wb/cm^2
and 1 Wb = kg*m^2/C*s
Then we have:
4*10^(-6) Wb/cm^2 = 4*10^(-6) kg*m^2/C*s*cm^2
= 4*10^(-2) kg/C*s*
Now we replace that to get:
[tex]r = \frac{1}{4*10^{(-2)} kg/C*s}*1.567*10^{-7} kg*km/s*C} = 0.0000039 km[/tex]
If you want to learn more about magnetic fields, you can read:
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