1 An electron moving at right angles to a 014 Tmagnetic fiel

1. An electron moving at right angles to a 0.14 Tmagnetic field experiences an acceleration of 6.4×1015 m/s2
.What is the electron\'s speed?
Express your answer using two significant figures.
v=m/s
Part B
By how much does its speed change in 1ns(109s)?
v= m/s

2.The magnitude of Earth\'s magnetic field is about 0.5 G near Earth\'s surface.
Part A
What\'s the maximum possible magnetic force on an electron with kinetic energy of 1keV?
Express your answer using two significant figures.

Fm= N

Part B
Compare with the gravitational force on the electron.
Express your answer using two significant figures.
Fg/Fm=

Solution

Magnetic field B = 0.14 T

Acceleration a = 6.4×10¹⁵ m/s²

Force F = ma

Where m = mass of electron = 9.1 x10 ^-31 kg

Substittue values you get F = ( 9.1 x10 ^-31 kg)( 6.4×10¹⁵ m/s² )

                                         = 5.824 x10 ^-15 N

We know F = Bvq           Where q = charge of electron = 1.6 x10 ^-19 C

From this electron\'s speed v = F/(Bq)

                                           = (5.824x10 ^-15 )/(0.14 x1.6x10 ^-19 )

                                           = 260x10 ³ m/s

                                           = 2.6 x10 ⁵ m/s
(B). we know F = q(vXB)

i.e., F is perpendicular to both velocity v and magnetic field B.

The magnetic force F on a charged particle changes its direction but not in its speed(i.e., in magnitude)