A proton moves in a circular path of the same radius as a cosmic ray electron moving at 6.5 × 10^6 m/s perpendicular to the Earth’s magnetic field at an altitude where the field strength is 1.0 × 10^(-5) T.
What is the radius of the circular path the electron follows?

3.7 m

Explanation:

We are given that

Mass of electron,m=

Magnetic filed,B=

Charge on electron,q=

We have to find the radius of circular path the electron follows.

We know that

Radius of circular path,r=

Using the formula

Hence, the radius of circular path followed by electron=3.7 m

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be