# Chapter 15 Motion of a charged particle¶

## Example 15.1 Page no 254¶

In :
#given
E=5000                                        #Intensity of electric field in N/C
d=0.02                                        #Distance in m
e=(1.6*10**-19)                               #Charge of the electron in C
m=(9.1*10**-31)                               #Mass of the electron in kg

#Calculations
import math
v=math.sqrt(2*e*E*d/m)/10**6

#Output
print"Speed of the electron is ",round(v,2),"*10**6 m/s"

Speed of the electron is  5.93 *10**6 m/s


## Example 15.2 Page no 255¶

In :
#given
v=(5*10**6)                                      #Velocity of the electron in m/s
E=2000                                           #Intensity of electric field in N/C
d=0.06                                           #Distance in m
e=(1.6*10**-19)                                  #Charge of the electron in C
m=(9.1*10**-31)                                  #Mass of the electron in kg

#Calculations
y=((-e*E*d**2)/(2*m*v**2))*100

#Output
print"Vertical displacement of the electron when it just leaves the electric field is ",round(y,2),"cm"

Vertical displacement of the electron when it just leaves the electric field is  -2.53 cm


## Example 15.3 Page no 255¶

In :
#given
v=(4*10**5)                                    #Velocity of the positively charged particle in m/s
E=300                                          #Intensity of electric field in N/C
e=(1.6*10**-19)                                #Charge of the positively charged particle in C
m=(1.67*10**-27)                               #Mass of the positively charged particle in kg
q=35                                           #Angle made by the particle in degrees

#Calculations
import math
t=((v*math.sin(q*3.14/180.0)*m)/(e*E))/10**-6

#Output
print"Time required by the particle to reach the maximum height in the electric field is ",round(t,2),"micro s"

Time required by the particle to reach the maximum height in the electric field is  7.98 micro s


## Example 15.4 Page no 255¶

In :
#given
r=0.3                                    #Radius of circular orbit in m
B=0.38                                   #Magnetic field strength in T
e=(1.6*10**-19)                          #Charge of the proton in C
m=(1.672*10**-27)                        #Mass of the proton in kg

#Calculations
v=((e*B*r)/m)/10**6

#Output
print"Orbital speed of the proton is ",round(v,0),"*10**6 m/s"

Orbital speed of the proton is  11.0 *10**6 m/s


## Example 15.5 Page no 255¶

In :
#given
e=(1.6*10**-19)                                      #Charge of the proton in C
m=(1.67*10**-27)                                     #Mass of the proton in kg
B=0.8                                                #Magnetic field strength in T
v=(4*10**6,3*10**6)                                  #Velocity of charged particle in vxi+vyj form in m/s

#Calculations
p=(v*2*3.14*m)/(e*B)
R=(m*v)/(e*B)

#Output
print"The pitch of the helix is ",round(p,3),"m"
print"Radius of the trajectory is ",round(R,3),"m"

The pitch of the helix is  0.328 m
Radius of the trajectory is  0.039 m