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

Speed of the proton is given as

`v = 2.1 * 10^6 m/s`

Step-by-step explanation:

As we know that force on moving charge is given as

`F = q(v* 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 * 10^(-12) = (1.6 * 10^(-19))(v)(1.9) sin90`

`v = 2.1 * 10^6 m/s`