Question

A proton having an initial velvocity of 20.0i Mm/s enters a uniform magnetic field of magnitude 0.300 T with a direction perpendicular to the proton's velocity. It leaves the field-filled region with velocity -20.0j Mm/s. Determine(d) the time interval during which the proton is in the field.

Answer 1

The time interval for which the proton remains in the field is -

Δt = $\frac{\pi R}{40}$.

We have a proton entering a uniform magnetic field which is in a direction perpendicular to the proton's velocity.

We have to determine time interval during which the proton is in the field.

What is the magnitude of force on the charged particle moving in a uniform magnetic field?

The magnitude of force on the charged particle moving in a uniform magnetic field is given by -

F = qvB sinθ



According to the question, we have -

Entering Velocity (v) = 20 i  m/s

Magnetic field intensity (B) = 0.3 T

Leaving velocity (u) = - 20 j  m/s

Now -

The entering and leaving velocity vectors have 90 degrees difference between them. Therefore, only a quarter of distance of the complete circular path of radius 'R' is traced by the proton. Therefore -

d = $\frac{2\pi r}{4}$ = $\frac{\pi R}{2}$

Since, the radius of circular path is not given, we will assume it R.

Therefore, time for which proton remained in the field is -

t = $\frac{\pi R}{2v} = \frac{\pi R}{40}$

Hence, the time interval for which the proton remains in the field is -

Δt = $\frac{\pi R}{40}$

To solve more questions on Force on charged particle, visit the link below-

brainly.com/question/14597200

#SPJ4