Chapter 8: Magnetic Forces, Materials and Devices

Example 8.1, Page number: 308

In [1]:
import scipy
from numpy import *

#Variable Declaration

m=2                     #mass in kg
q=3                     #charge in C
v=array([4,0,3])        #initial velocity in m/s
E=array([12,10,0])      #electric field in V/m
t=1                     #time in sec

#Calculations

a=q*E/m                                 #acceleration in m/s^2 after 1 sec
u=array([18*t+4,15*t,3])                #velocity in m/s after 1 sec
modofu=scipy.sqrt(dot(u,u))
KE=0.5*m*(modofu)**2                    #kinetic energy in J at t=1 sec
s=array([9*t**2+4*t+1,7.5*t**2-2,3*t])  #position after 1 sec in m

#Results

print 'At time t=1 sec,'
print' The acceleration of the particle =',a,'m/s^2'
print 'Its velocity =',u,'m/s' 
print 'Its kinetic energy =',KE,'J'
print 'Its position =',s,'m'

At time t=1 sec,
 The acceleration of the particle = [18 15  0] m/s^2
Its velocity = [22 15  3] m/s
Its kinetic energy = 718.0 J
Its position = [ 14.    5.5   3. ] m

Example 8.6, Page number: 322

In [14]:
 

import scipy

#Variable Declaration

ar=array([1,0,0])                    #Unit vector along radial direction
ath=array([0,1,0])                   #Unit vector along theta direction
aph=array([0,0,1])                   #Unit vector along phi direction 
x=4
y=-3
z=10
muo=4*scipy.pi*10**-7                #permeability of free space
m1=5                                 #magnetic moment in A/m^2

#Calculations

r=scipy.sqrt(x**2+y**2+z**2)
p=scipy.sqrt(x**2+y**2)
sinphi=y/p
cosphi=x/p
sintheta=1/scipy.sqrt(5)
costheta=2/scipy.sqrt(5)
B1=muo*m1*(2*costheta*ar+sintheta*ath)/(4*scipy.pi*r**3)
m2=3*(sintheta*sinphi*ar+costheta*sinphi*ath+cosphi*aph)
T2=cross(m2,B1)*10**9
T2x=round(dot(T2,ar),3)
T2y=round(dot(T2,ath),3)
T2z=round(dot(T2,aph),3)
T2r=array([T2x,T2y,T2z])      #torque in nNm

#Result

print 'Torque T2 =',T2r,'nNm'

Torque T2 = [-0.384  1.536  0.902]

Example 8.7, Page number: 330

In [4]:
import scipy

#Variable Declaration

muo=4*scipy.pi*10**-7               #permeability of free space
mur=2                               #relative permeability
ax=array([1,0,0])                   #Unit vector along x direction
ay=array([0,1,0])                   #Unit vector along y direction
az=array([0,0,1])                   #Unit vector along z direction


#Calculations

J=(-5-10)*10**-6/(4*scipy.pi*10**-7*2.5)          #in kA/m^2
Jb=1.5*J                                          #in kA/m^2
MbyB=(1.5)*10**4/(4*scipy.pi*2.5)       
Mv=MbyB*10*10**-3*ax+MbyB*5*10**-3*ay
Kb=cross(az,Mv)

#Results

print 'J =',round(J,3),'kA/m^2'
print 'Jb =',round(Jb,3),'kA/m^2'
print 'M =(',round(dot(Mv,ax),3),'y,',round(dot(Mv,ay),3),'x, 0) kA/m'
print 'Kb =(',round(dot(Kb,ax),3),'x,',round(dot(Kb,ay),3),'y, 0) kA/m' 
         

 

J = -4.775 kA/m^2
Jb = -7.162 kA/m^2
M =( 4.775 y, 2.387 x, 0) kA/m
Kb =( -2.387 x, 4.775 y, 0) kA/m

Example 8.8, Page number: 332

In [2]:
 
import scipy
from numpy import *

#Variable Declaration

ax=array([1,0,0])                   #Unit vector along x direction
ay=array([0,1,0])                   #Unit vector along y direction
az=array([0,0,1])                   #Unit vector along z direction 
H1=array([-2,6,4])                  #in A/m
mu0=4*scipy.pi*10**-7               #permeability of free space
mur1=5                              #relative permeabililty in region 1
mur2=2                              #relative permeabililty in region 2
an=array([-1,1,0])/scipy.sqrt(2)

#Calculatios

mu1=mu0*mur1
mu2=mu0*mur2
M1=(mur1-1)*H1                      # magnetisation in region 1 in A/m
B1=mu1*H1*10**6                     # field in micro Wb/m^2
B1x=round(dot(B1,ax),2)             # x component of B1
B1y=round(dot(B1,ay),1)             # y component of B1
B1z=round(dot(B1,az),2)             # z component of B1
B1r=array([B1x,B1y,B1z])            # B1 rounded to 2 decimal places
H1n=dot(H1,an)*an                   
H1t=H1-H1n
H2t=H1t                             # using transverse boundary condition
H2n=(mu1/mu2)*H1n                   # using normal boundary condition
H2=H2t+H2n                          # in A/m
B2=mu2*H2*10**6                     # field in micro Wb/m^2
B2x=round(dot(B2,ax),2)             # x component of B2
B2y=round(dot(B2,ay),2)             # y component of B2
B2z=round(dot(B2,az),2)             # z component of B2
B2r=array([B2x,B2y,B2z])            # B2 rounded to 2 decimal places

#Results

print 'M1= ',M1,'A/m'
print 'B1= ',B1r,'micro Wb/m^2'
print 'H2= ',H2,'A/m'
print 'B2= ',B2r,'micro Wb/m^2'

M1=  [-8 24 16] A/m
B1=  [-12.57  37.7   25.13] micro Wb/m^2
H2=  [ -8.  12.   4.] A/m
B2=  [-20.11  30.16  10.05] micro Wb/m^2

Example 8.14, Page number: 350

In [3]:
 
import scipy
from numpy import *

#Variable declaration

p=10*10**-2             #in m
a=1*10**-2              #in m
Ur=1000                 #relative permeability
Uo=4*scipy.pi*10**-7    #permeability of free space
n=200                   #number of turns
phi=0.5*10**-3          #flux in the core in Wb
U=Uo*Ur                 #permeability of steel core

#Calculation

I=phi*2*scipy.pi*p/(U*n*scipy.pi*a*a)   #current in A

#Result

print 'The current that will produce a flux of 0.5 mWb =',round(I,3),'A'

The current that will produce a flux of 0.5 mWb = 3.979 A

Example 8.15, Page number: 351

In [4]:
 

import scipy
from numpy import *

#Variable Declaration

Uo=4*scipy.pi*10**-7    #permeability of free space
Ur=50                   #relative permeability of coil
l1=30*10**-2
s=10*10**-4 
l3=9*10**-2
la=1*10**-2 
B=1.5                   #flux density in Wb/m^2
N=400                   #number of turns

#Calculations

R1=l1/(Uo*Ur*s)
R2=R1
R3=l3/(Uo*Ur*s)
Ra=la/(Uo*s)
R=R1*R2/(R1+R2)
Req=R3+Ra+R
I=B*s*Req/N     #current in A

#Result

print 'The current required =',round(I,3),'A'

The current required = 44.165 A

Example 8.16, Page number: 353

In [5]:
 

import scipy
from numpy import *

#Variable Declaration

m=400                       #mass in kg
g=9.8                       #acceleration due to gravity in m/s^2
Ur=3000                     #relative permeability of the iron yoke
Uo=4*scipy.pi*10**-7        #permeability of free space
S=40*10**-4                 #cross sectional area of iron yoke in m^2
la=1*10**-4                 #air gaps in m
li=50*10**-2                #mean length of yoke in m
I=1                         #excitation current in A 

#Calculations

B=scipy.sqrt(m*g*Uo/S)      #field in Wb/m^2
Ra=2*la/(Uo*S) 
Ri=li/(Uo*Ur*S) 
N=(Ra+Ri)/(Ra*Uo)*B*la      #number of turns

#Result

print 'The nmber of turns in the coil when the excitation current is 1 A ='
print round(N,0)

The nmber of turns in the coil when the excitation current is 1 A =
162.0