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The radius of curvature of the path of a 14.4 MeV proton in a 6.00 T magnetic field is 0.091 m.
The radius of curvature can be calculated with Lorentz force:
[tex] F = q(E + v\times B) [/tex]
Since there is no electric field (E = 0) and the Lorentz force is equal to the centripetal force, we have:
[tex] ma_{c} = q(v\times B) [/tex]
Replacing [tex]a_{c}[/tex] for [tex]frac{v^{2}}{r}[/tex] in the above equation, we get:
[tex] \frac{mv^{2}}{r} = q(v\times B) [/tex]
Where:
m: is the mass of the proton = 1.67x10⁻²⁷ kg
q: is the charge of the proton = 1.602x10⁻¹⁹ C
r: is the radius of curvature =?
v: is the tangential velocity
B: is the magnetic field = 6.00 T
The magnetic field is perpendicular to the charged particles, so:
[tex] \frac{mv^{2}}{r} = q[vBsin(90)] [/tex]
[tex] \frac{mv^{2}}{r} = qvB [/tex]
[tex] r = \frac{mv}{qB} [/tex]
The tangential velocity can be calculated from the energy:
[tex] E = \frac{1}{2}mv^{2} [/tex]
[tex] v = \sqrt{\frac{2E}{m}} = \sqrt{\frac{2*14.4 \cdot 10^{6} eV*\frac{1.602 \cdot 10^{-19} J}{1 eV}}{1.67 \cdot 10^{-27} kg}} = 5.26 \cdot 10^{7} m/s [/tex]
Finally, the radius is:
[tex] r = \frac{mv}{qB} = \frac{1.67 \cdot 10^{-27} kg*5.26 \cdot 10^{7} m/s}{1.602 \cdot 10^{-19} C*6.00 T} = 0.091 m [/tex]
Therefore, the radius of curvature of the path is 0.091 m.
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