Answer:
The acceleration of the proton is [tex]4.72\times 10^{14}\ m/s^2[/tex].
Explanation:
Given that,
Speed of a proton, [tex]v=2.9\times 10^7\ m/s[/tex]
The proton moves with this speed at right angles to a magnetic field with a magnitude of 0.17 T, B = 0.17 T
Charge on the proton, [tex]q=1.6\times 10^{-19}\ C[/tex]
The proton moves in the magnetic field such that the the magnetic force is balanced by the force due to its motion as :
[tex]ma=qvB\sin\theta\\\\ma=qvB\\\\a=\dfrac{qvB}{m}\\\\a=\dfrac{1.6\times 10^{-19}\times 2.9\times 10^7\times 0.17}{1.67\times 10^{-27}}\\\\a=4.72\times 10^{14}\ m/s^2[/tex]
So, the acceleration of the proton is [tex]4.72\times 10^{14}\ m/s^2[/tex]. Hence, this is the required solution.
