Answer:
Magnetic field, [tex]B=4.16\times 10^{-5}\ T[/tex]
Explanation:
It is given that,
Velocity of proton, [tex]v=3.5\times 10^7\ m/s[/tex]
Magnetic force, [tex]F=1.65\times 10^{-16}\ N[/tex]
Charge of proton, [tex]q=1.6\times 10^{-19}\ C[/tex]
We need to find the strength of the magnetic field if there is a 45° angle between it and the proton's velocity. The formula for magnetic force is given by :
[tex]F=qvB\ sin\theta[/tex]
[tex]B=\dfrac{F}{qv\ sin\theta}[/tex]
[tex]B=\dfrac{1.65\times 10^{-16}}{1.6\times 10^{-19}\times 3.5\times 10^7\times sin(45)}[/tex]
B = 0.0000416 T
[tex]B=4.16\times 10^{-5}\ T[/tex]
Hence, this is the required solution.
