A pendulum is raised 1.5 cm and allowed to fall. if air resistance is negligible, how high will the pendulum rise on the other s
2 answers:
You might be interested in
The time interval for which the proton remains in the field is -
Δt = .
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.
<h3>What is the magnitude of force on the charged particle moving in a uniform magnetic field?</h3>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 = =
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 =
Hence, the time interval for which the proton remains in the field is -
Δt =
To solve more questions on Force on charged particle, visit the link below-
#SPJ4