Question

A proton is initially moving west at a speed of 1.10 106 m/s in a uniform magnetic field of magnitude 0.281 T directed vertically upward. Describe in detail the proton's trajectory, including its shape and orientation.

Answer 1

A proton moving west in a uniform magnetic field directed upward will follow a circular path that lies in a horizontal plane due to the perpendicular magnetic force it experiences as per the right-hand rule.

Step-by-step explanation:

When a proton is moving west in a uniform magnetic field that is directed vertically upward, the proton will experience a magnetic force due to its motion through the magnetic field. This magnetic force will be perpendicular to both the direction of the proton's velocity and the magnetic field (right-hand rule). As the force is always perpendicular to the velocity, the proton will begin to move in a circular path with a radius determined by the balance of this magnetic force and the proton's momentum.

For a proton moving westward in a uniform magnetic field directed upwards, the magnetic force will act in a direction that is towards the north or south depending on the orientation of the magnetic field relative to the proton's velocity. This circular motion will lie in a horizontal plane. The shape of the trajectory is a circle, and the orientation of this circle is in the horizontal plane perpendicular to the magnetic field.

Answer 2

so here it will move in circle with radius 4.06 cm

Step-by-step explanation:

As we know that proton is moving towards west while the magnetic field is vertically upwards

So here the force on the proton must be perpendicular to the velocity

So here we have

`F = q(\vec v * \vec B)`

so here we have

`F = qvB sin90`

since force is perpendicular to the velocity so here it must be centripetal force

here we have

`(mv^2)/(R) = qvB`

so we have

`R = (mv)/(qB)`

`R = ((1.66 * 10^(-27))(1.10 * 10^6))/((1.6 * 10^(-19))(0.281))`

`R = 4.06 cm`

so here it will move in circle with radius 4.06 cm