G An electron enters a region of space containing a uniform 1.63 × 10 − 5 T magnetic field. Its speed is 121 m/s and it enters perpendicularly to the field. Under these conditions, the electron undergoes circular motion. Find the radius r of the electron's path and the frequency f of the motion.

i. The radius 'r' of the electron's path is 4.23 × m.

ii. The frequency 'f' of the motion is 455.44 KHz.

Explanation:

The radius 'r' of the electron's path is called a gyroradius. Gyroradius is the radius of the circular motion of a charged particle in the presence of a uniform magnetic field.

                 r =

Where: B is the strength magnetic field, q is the charge, v is its velocity and m is the mass of the particle.

From the question, B = 1.63 × T, v = 121 m/s, Θ = (since it enters perpendicularly to the field), q = e  = 1.6 × C and m = 9.11 × Kg.

Thus,

         r = ÷ sinΘ

But,  sinΘ =  sin = 1.

So that;

          r =

            = (9.11 × × 121) ÷ (1.6 ×  × 1.63 × )

            = 1.10231 ×   ÷ 2.608 ×

            = 4.2266 ×

            = 4.23 × m

The radius 'r' of the electron's path is 4.23 × m.

B. The frequency 'f' of the motion is called cyclotron frequency;

           f =

             =  (1.6 ×  × 1.63 × ) ÷ (2 × × 9.11 × )

             =  2.608 × ÷  5.7263 ×

             = 455442.4323

          f  = 455.44 KHz

The frequency 'f' of the motion is 455.44 KHz.

real, smaller and upside-down.   so the answer should be b.

temp rises, and expansion