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Chapter V
Magnetostatics
Recommended problems: 4.5, 4.6, 4.8, 4.9, 4.10, 4.11, 4.12, 4.15, 4.18,
4.19, 4.20, 4.21, 4.22, 4.24, 4.26, 4.27, 4.31, 4.32, 4.33, 4.34, 4.36
The Lorentz Force Law
It is known that any stationary charge produces an e.f. If a charge q is

placed in an e.f. E this charge experiences an e. force according to


F q E
Now a moving charge produces a m.f, and any moving charge in a
magnetic field experiences a m. force according to.

 
Fm  q v  B
Lorentz force law

Where B is the magnetic field strength. It is clear that



Fm  v & B

 ds
Now since v 
dt

 
F
 m  ds  0

The magnetic force doesn't do any work in a moving charge.


If E and B are coexist 
  

 
F  Fe  Fm  q E  q v  B
Motion of a Charged Particle in M. Field
Since Wm  0  K  0  The speed is constant  the acceleration is

due to the change in direction only Fm is centripetal, that is the motion
of the charge is circular.
Let us prove this concept more quantitatively.
Let

B  Bkˆ
with B is constant
Now


d 2r
 
Fm  q v  B  m 2
dt
But
iˆ
ˆj
kˆ
 
v  B  x
y
z  By iˆ  Bx ˆj
0
0
B
mxi  y j  z k   qB y iˆ  x ˆj 


mx  qBy
(1)
my   qBx
(2)
mz  0
(3)
Integrating w.r.t. t 
mx  qBy  C1
(4)
my   qBx  C2
(5)
z  C3
(6)
Substituting for y from Eq. (5) into Eq. (1) 
2
 qB 
x     x  C
m

x   2 x   2 a
where  
(7)
qB

m
x  a  R cost o 
(8)
With a, R, ando are constants.
Differentiating Eq. (8) and substituting into Eq. (4) 
  R sin t o y  C
y  b  R sin t  o 

(9)
Eliminating t from Eqs (8) & (9) 
x  a 2   y  b2  R 2
 The path of the motion is a circle of radius R and centered at (a,b).
(10)
Since z  Constant  The path is spiral with its axis in the direction of

B
Now from Eq.(8) we have
x   R sin t o 
(11)
And from Eq. (9) we have
y   R  cost o 
(12)
Eliminating t from Eqs (11) & (12) 

x 2  y 2  R 2 w2
R
1

If z  0 
x 2  y 2
R
v
mv
 qB

Example 5.2: Cycloid Motion. A particle of mass m and charge q is starts



from rest at the origin in the presence of both E and B such that E  E kˆ

and B  B iˆ . Describe the motion of the particle.
Since the particle starts from rest  x  0 (since Fx  0 )
Now
ˆ ˆj kˆ
i
 
v  B  x y z  Bz ˆj  By kˆ
B 0 0
  

 
Applying F  Fe  Fm  q E  q v  B

m r  qE kˆ  qB z ˆj  y kˆ 



m y  qBz
m z  qE  qBy 
&
y   z
z
(1)
qE
  y
m
(2)
Integrating Eq. (1) 
y   z  C1   z
(since y  0 at z  0 )
Substituting back in Eq. (2) 
z 
qE
2 z 
m
z  2 z 
qE

m
z  R  A cost  o 
(3)
Now integrating Eq. (2) we get
z 
qE
qE
t   y  C2  t   y
m
m
(4)
(since z  0 at y  0 and t  0 )
Differentiating Eq. (3) and equating the result with that of Eq. (4) 
 A sin t  o  
qE
t  y
m
y  A sin t   o  

qE
t
m
(5)
Eliminating t from Eqs (3) & (5) 
 y  Rt 2  z  R2  R2
With R 
E
E
and the speed is v  R 
B
B
(6)
Current
A moving charge in one direction produces current I with
I
 
dq
 neA  vd
dt
With A is the cross sectional area of the wire, n is the No. of free charges
per unit volume, and vd is the drift velocity. The magnetic force on
current-carrying wire is

Fm 
 
 
 
 v  Bdq   v  BIdt   I dl  B


Example 5.3: A rectangular loop of wire,
B
supporting a mass m, hangs vertically with one
end in a uniform m.f. B. Find I such that m be in
equilibrium.
I
a


Fm  Fg  0 


Fm   Fg  mg ˆj
 

Fm  I l  B  IaB (iˆ  kˆ)  IaB ˆj


The Current Density
m

I
mg
aB

The current density J is defined such that
 
I   J  dA
Or


J   vd
Now

 
 
 
Fm   v  B dq    v  B d   J  B d
The Biot-Savart Law
The m.f at a point due to a wire of steady current I is given by

 o I dl   
B
4   3

with dl  is an element along the wire
  
  r  r  is a vector from the source to the point, and
 o  4  10 7 N / A2 is the magnetic permeability of free space.
Example 5.5: Find the m.f a distance s from a long straight wire carrying
a steady current I.

dl   dz kˆ

r  s sˆ
&
P

r   z kˆ

  
  r  r   s sˆ  z kˆ

  o I dl     o I dzkˆ  s sˆ  z kˆ
B

4   3
4 
3


B 
 o Is
4

Knowing that

s
dz
2
z
dx
a x 
2
3
2
 o I 
z
B 
4s  s 2  z 2


If
2

3

s
I
z

dz
2
x

2
a
2
l
a x
2
2  I
  o
3
2 
4s
 l1

2

 l

l1
2



2
2
2
2
 l2  s
l1  s 
l1  l2   
 o Il
B  lim
l  2s
l22  s 2

o I
2s
To find the force between two parallel wires
we have for the force on wire 1 due to wire 2
 

F12  I1 l1  B2

I1
d
I2

But

 I
B2  o 2 kˆ is the m.f due to I2 at the position of I1
2d


 II
F12  o 1 2 l  ˆj 
2d
The force is attractive if the currents are in the same direction and
repulsive if the currents are in opposite directions.
Example 5.6: Find the m.f due a distance z from the center of a circular
loop of radius R which carries a steady current I.

dl   Rd ˆ  Rd  sin  iˆ  cos ˆj 

r  z kˆ &

r   R sˆ  Rcos  iˆ  sin  ˆj  
P
z
  
  r  r   z kˆ  R cos iˆ  R sin  ˆj
ˆj
kˆ
R cos  d
0
 R sin 
z
iˆ
 
 dl      R sin  d
 R cos 

dl'

 Rz cos iˆ  Rz sin  ˆj  R 2 kˆ d


 o I dl    o I 2 Rz cos iˆ  Rz sin  ˆj  R 2 kˆ d
B


3
3
4
4 0
2
2 2

R z
 Bx  B y  0 and Bz 



 o IR 2
2R z
2
2

3
2


Note that at the origin of the loop ( z  0 ) we get Bz 

Divergence and Curl of B

 o I dl   

B

3
4

 
  o I   dl   
 B
4 
3

o I
2R

Knowing that
  
     

  A  B    A  B    B  A


   o I   dl     o I     
 
 dl 
 B




3
3
4
4 

 

 
 1

But 3    and     0 

 





 
 B0


 o I dl    o J r  
Now B 
 
d 
3
4   3
4

 
  o    

 B 
   J r   3  d 

4
 

        
Using the bac-cab rule A  B  C  BA  C  C A  B  we obtain
        
  



   J r   3   J    3   3   J 
     

  
 

Knowing that   J r   0 and   3  4  

 
 
 
 
  B  o  J r  r  r  d   o J r 

 
 
  B  o J r 
Ampere's Law
Using Stoke's theorem we have
  
 
   B  dS   B  dl
 
 
But   B   o J r 
 
 
 B  dl  o  J  dS


 
 B  dl  o I enc
Where I enc is the total current crossing the surface enclosed by the closed
loop.
Example 5.7: Find the m.f a distance s from a long straight wire carrying
a steady current I.
P
s
I