A proton traveling to the right along the x-axis enters a region where there is a mag-netic field of magnitude 1.9 T directed upwardalong they-axis.The charge on a proton is 1.60×10−19C.If the proton experiences a force of 0.64×10−12N, find its speed.Answer in units of m/s.
Question
Answer:
Speed of the proton is given as
[tex]v = 2.1 \times 10^6 m/s[/tex]
Explanation:
As we know that force on moving charge is given as
[tex]F = q(v\times B)[/tex]
here we know that charge is moving perpendicular to the field
So we will have
[tex]F = qvB sin90[/tex]
here we will have
[tex]0.64 \times 10^{-12} = (1.6 \times 10^{-19})(v)(1.9) sin90[/tex]
[tex]v = 2.1 \times 10^6 m/s[/tex]