Answer:
Magnetic force, [tex]F=1.12\times 10^{-13}\ N[/tex]
Explanation:
It is given that,
Velocity of proton, [tex]v=1.8\times 10^6\ m/s[/tex]
Angle between velocity and the magnetic field, θ = 53°
Magnetic field, B = 0.49 T
The mass of proton, [tex]m=1.672\times 10^{-27}\ kg[/tex]
The charge on proton, [tex]q=1.6\times 10^{-19}\ C[/tex]
The magnitude of magnetic force is given by :
[tex]F=qvB\ sin\theta[/tex]
[tex]F=1.6\times 10^{-19}\ C\times 1.8\times 10^6\ m/s\times 0.49\ Tsin(53)[/tex]
[tex]F=1.12\times 10^{-13}\ N[/tex]
So, the magnitude of the magnetic force on the proton is [tex]1.12\times 10^{-13}\ N[/tex]. Hence, this is the required solution.
