Question

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.

Answer 1

Answer:

Speed of the proton is given as

$v = 2.1 \times 10^6 m/s$

Explanation:

As we know that force on moving charge is given as

$F = q(v\times B)$

here we know that charge is moving perpendicular to the field

So we will have

$F = qvB sin90$

here we will have

$0.64 \times 10^{-12} = (1.6 \times 10^{-19})(v)(1.9) sin90$

$v = 2.1 \times 10^6 m/s$