Explanation:
It is given that,
Charge on proton, [tex]q=1.6\times 10^{-19}\ C[/tex]
Magnetic field, B = 2 T
Speed of proton, [tex]v=2.5\times 10^6\ m/s[/tex]
To find,
The force experienced by proton and its acceleration.
Solution,
The force experienced by a proton is given by :
[tex]F=qvB[/tex]
[tex]F=1.6\times 10^{-19}\times 2.5\times 10^6\times 2[/tex]
[tex]F=8\times 10^{-13}\ N[/tex]
We know that the product of mass and acceleration is equal to force. If a is the acceleration of proton. So,
[tex]a=\dfrac{F}{m}[/tex]
[tex]a=\dfrac{8\times 10^{-13}}{1.67\times 10^{-27}}[/tex]
[tex]a=(4.79\times 10^{14})\ m/s^2[/tex]
