A uniform magnetic field of magnitude 025 T is directed alon

A uniform magnetic field of magnitude 0.25 T is directed along the positive x-axis. A positron moving at 2.5 time sign 10^6 m/s enters the field along a direction that makes an angel of 70 degree with the x-axis. The motion of the particle is expected to be a helix. Calculate: The pitch p and the radius of the trajectory.

Solution

For a charge q entering in to a magnetic field B at an angle of , the force acting on it is given as qvBSin

Now this force provides for the centripetal force that causes the charged particle to move in the circular path. However, the particle will also have a component of the velocity in the direction of the magnetic field and will be given as VSin

So by the time the particle completes one revolution of the circle, it will cover a horizontal distance and that distance is called pitch.

Now, for the centripetal force we can write:

Mv²/R = qvBSin

or, R = Mv / qBSin

Now time to complete one complete circle = 2R / v = 2M / qBSin

Therefore the distance travelled along the direction of the field during the time the particle completes one circle = VSin x Time

Hence, pitch = VSin x 2M / qBSin = 2vM / qB

and the radius of trajectory = Mv / qBSin

Putting in the values we get: Pitch = 2 x 2.5 x 10^6 x 9.109 x 10^-31 / 1.60217662 × 10^-19 x 0.25

or, Pitch = 113.7078 x 10^-6 m

Radius = Mv / qBSin = 9.109 x 10^-31 x 2.5 x 10^6 /  1.60217662 × 10^-19 x 0.25 Sin70

or, Radius = 60.502 x 10^-6 m