1. M
MA
AG
GN
NE
ET
TI
IC
CS
S
Stationary charge carries only electric field where moving charge carries electric field as well as
magnetic field.
B
BI
IO
OT
T S
SA
AV
VA
AR
RT
T’
’S
S L
LA
AW
W :
:
The magnetic field due to a current element Idl


at a point is given by
the expression
0
2
sin
4
Idl
dB
r
 

 where
idl 


current element, its direction same as that of the current
r 

position vector of point P w.r.t. current element
  angle between current element and position vector
i

P
dl r
where 0
  Permeability of free space.
and 7 2
0 4 10 T mA
   
 
2. Magnetic field due to current carrying wire
0
2
. sin
4
i d
B dB
r
 

 
 

S.I. unit of B is Tesla
3. Vector form of magnetic field due to a current carrying wire is
0
3
( )
4
i d r
B
r



 

 
 
4. Direction of magnetic field can be found by the rules of vector product.
A
AP
PP
PL
LI
IC
CA
AT
TI
IO
ON
N O
OF
F B
BI
IO
OT
T S
SA
AV
VA
AR
RT
T’
’S
S L
LA
AW
W :
:
1. Magnetic field at a point on the line of current :
If a point lies in the line of current carrying element then magnetic
field at this point is always zero.
B=0
Q
I
P
B=0
2. Magnetic field due to a straight current carrying wire
(i) Of finite length : Suppose A straight current carrying wire AB, carrying current I, lies in
the plane of the paper. As shown in the figure, P is a point at a perpendicular distance R
from conductor, where magnetic field is to be determined
According to Biot – Savart’s Law, the field at P due to a current element Idl


is
0 0
2
sin sin 1
4 4
I
Id d
dB
r r r
 
 
 
 
   
 
 
Now from the figure sin
d rd
 

 and cos
R
r
 
0 cos
.
4
I
dB d
R
 


 
Hence magnetic field due to the whole conductor
B
I
P
r
d
1
2

A
R
Idl

2. 1 1
2
0 0
1 2
cos . [sin sin ]
4 4
I I
B dB d
R R
 
 
 
   
 

   
 
2
-
0
1 2
[sin sin ]
4
I
B
R

 

  
(ii) For the conductor of infinite length 1 2
2

 
 
0
2
I
B
R


 
(iii) On perpendicular bisector of finite length :
a  length of the wire
d  perpendicular distance of the field point then
1 2 2 2 2
2
2
sin sin
4
2
a
a
a d
a
d
 
  

 

 
 
 Magnetic field 0
2 2
2
4 4
I a
B
R a d




P
1
2
a
d
(iv) At point exactly in front of one end of semi-infinite wire :
Here 1 0
  and 2
2

 
0
sin0 sin
4 2
I
B
R
 

 
  
 
 
0
4
I
R



P
2

(v) At a point not exactly in front of the end of a semi infinite wire :
Here 1
 
 and 2
2

 
0
(sin sin )
4 2
I
B
R
 


  
0
(1 sin )
4
I
R



 
P
2


Application : Find magnetic field at point P shown in figure, the point P
is on the bisector of angle between the wire.
Solution : Assume ,
PO x
 so sin
2
PQ x 

Magnetic field B at P due to either segment of wire is,
B 0
1 cos
2
4 sin
2
I
x
 


 
 
 
 
Net magnetic field at P is
P

I
Q
O x
I

3. 0
2 sin
2
net
I
B
x



  (1 sin )
2


3. Magnetic field due to circular arc at the center (Subtending an angle  at the center) :
Consider a current element that subtends an angle  as shown in
the figure. Magnetic field due to this element.
Thus, 0 0
2
.sin90
4 4
I I
Rd
dB d
R R
 


 

  
I
d Rd

R
0 0
4 4
O
I I
B d
R R

  

 
  

(ii) Magnetic field at the centre of a loop:
Here the loop makes an angle 2
 
 at the center
0
2
I
B
R

 
for N turns, 0 0
2 2
I NI
B N B
R R
 
 
  
 
 
R
I
Application : Find magnetic field at O, by the system of current
carrying wire.
Solution: As in figure O AB BCD DE
B B B B
  
   
0 0 0
ˆ ˆ
( ) ( )
4 4 4
O
I I I
B j i
R R R
  
 
   
     
   
   
 B
y
O D
A
x
z

E
I
I
I
C
4. Magnetic field at any point on the axis of a circular current carrying coil :
Consider a circular conducting coil of radius R carrying current I.
The loop lies on yz plane and its axis lies on x axis. Let us derive
field at point P at a distance x from the center. Consider a small
element at  making an angle d at the center
Here 0
2 2
sin
4 ( )
Idl
dB
R x
 



As the loop lies perpendicular to the plane of paper and vector r

in the plane of the paper hence angle  between dl


and r

is 90º
y
x
z

dB
d

2 2
R x r
 
x
dl
0
2 2
4 ( )
Idl
dB
R x


 

Magnetic field dB

can be resolved into two components one sin
dB  parallel to the axis of the
loop and other cos
dB  perpendicular to the axis.

4. From the symmetry of the system it can be seen that diametrically opposite elements contribute to
cancel the perpendicular components whereas parallel component are added up.
sin
B dB 
  
Thus, 0 0
2 2
sin sin
4 4
I I
dl
B dl
r r
 
 
 
 
 
0 0
2 2 2 2 2
sin .2 .
4 2( )
I I R
B R
r R x R x
 
 

  
 
2
0
3
2 2 2
2
4 ( )
IR
B
R x
 

 

for single turn
and
2
0
3
2 2 2
2
4 ( )
NIR
B
R x
 



for N turns
N
No
ot
te
e :
:
(i) This magnetic field is directed along the axis of ring
(ii) The field strength is the maximum at the center (where x = 0).
Magnetic field at the center, 0
2
NI
B
R


(iii) At very large distance when
2
0
3
,
2
NIR
x R B
x

 
5. Magnetic field due to solenoid :
A solenoid is a long cylindrical helix, which is obtained by
winding closely a large number of turns of insulated copper wire
over a tube of cardboard or china clay. When electric current is
passed through it, a magnetic field is produced around and within
the solenoid.
If n be the number of turns per unit length of the solenoid than
number of turns in width .
dx ndx
 If i current is passing
through it then magnitude of magnetic induction at a point P on
its axis due to this element
2
0
3
2 2 2
( )
2( )
ndx iR
dB
R x



along the axis.
 Net magnetic induction
1
2
2
0
3
2 2 2
2( )
l
l
niR dx
B dB
R x


 

 
0 0
1 2
1 2
2 2 2 2
1 2
[cos cos ]
2 2
ni ni
l l
B
R l R l
 
 
 
 
    
 
 
 
or 0
[cos cos ]
2
ni
B

 
 
which is directed along the axis.
dx
P x
R
l1 l2
1 2
P
2 2
1
R l

2 2
2
R l

P


Case I :
For ideal solenoid (i.e. solenoid of infinite length) R L


5. 0
in
B ni

  (i.e. same everywhere)
0
out
B 
Case II:
Semi-infinitely long solenoid
0 0
cos0 cos
2
2 2
ni ni
B
 

 
   
 
Application : an infinitely large sheet carries current with linear current density i. Find the net magnetic
field at a point which is at perpendicular distance r from the sheet.
Solution : Let us consider a current carrying element idx
0
2 2
( )
2 ( )
p
idx
dB
r x


 

It has two components one parallel to the plane of the
sheet and other perpendicular to it.
cos
x
dB dB 
  and sin
y
dB dB 
 
0 0
2 2
2 ( ) 2
x x
idxr i
B dB
r x
 



    

 
and 0
2 2
0
2 ( )
y y
idx x
B dB
r x


 
 
  

 
0
2
l
B

  
r
P
dx
x
dB

x
2 2
r x


Application : A conducting ring of radius r having charge q is rotating with angular velocity  about its
axes. Find the magnetic field at the center of ring.
Solution : Current in ring
2
q
I



Magnetic field 0 0
2 2 2
I q
B
r r
  

 

0
4
q
B
r
 

 
r
q

L
LO
OR
RE
EN
NT
TZ
Z F
FO
OR
RC
CE
ES
S :
:
1. If a charge q is moving with velocity V

enters in a region in which electric field E

and magnetic
field B

both are present, it experiences force due to both fields simultaneously. force given by
The force experienced by the charged particle is given by the expression
( )
F q V B qE
  
   
here magnetic force ( )
m
F q V B
 
  
and electric force e
F qE

 

6. 2. The direction of magnetic force is same as V B

 
if charge is positive and opposite to V B

 
is q
if charge negative
3. If q is charge without sign and  is the angle between V

and B , then the magnitude of magnetic
force is
sin
F q V B 

C
CA
AS
SE
ES
S O
OF
F P
PR
RO
OJ
JE
EC
CT
TI
IO
ON
N :
:
Case I:
If velocity of charge particle V

is parallel to B

then
sin 0
F qv B o F
    q
m
V
Case II:
If velocity of charge particle V

is perpendicular to B

then charge
particle leaves the magnetic field tangentially.
sin90
F q V B qV B
  
m
q
F
V
Note :
As the magnetic force is always perpendicular to the direction of motion of particle it can never
do work on it. Thus kinetic energy of charge particle in magnetic field can never charge.
1. Because path of charge in magnetic field is circular
2
mV mV
qV B R
R qB
  
2. Angular velocity
( )
B
Rq
V m
R R
  
qB
m
 
Case III:
If velocity V

is at angle  to B

, then charge particle moves in helix
1. cos
v   constant
B
vsin
vcos

v
helix
2. Radius of helix
2
( sin )
( sin )
m V
q V B
R



sin
mV
R
qB

 

7. 3. Time period of one revolution
2
sin
R
T
V



sin
2
sin
mv
qB
T
V



 
 
 
 
2 m
T
qB

 
4. Pitch of helix, ( cos )( )
H
P V T


2 2 cos
( cos )
H
m mv
P V
qB qB
  

 
  
 
 
Application : From the surface of a round wire of radius a carrying a direct current I an electron escapes
with a velocity 0
v perpendicular to the surface. Find what will be maximum distance of
the electron from the axis of the wire before it turns back due to the action of the
magnetic field generated by the current.
Solution : here 0 ˆ
2
A
I
B k
a



and 0 ˆ
2
P
I
B k
x



assume velocity of e at p is ˆ ˆ
p x y
v v i v j
 
here 2 2 2 2 2 2
0 0
x y y x
v v v v v v
    
y xmax
v =v
y 0
v =0
x
P(x,y)
a
(a,o,o)
x
I A
v0
2 2
0
y x
v v v
   …(i)
The magnetic force on the electron at point P
0 ˆ
ˆ ˆ
( ) ( )
2
x y
I
F qV B e v i v j k
x


 
      
 
  
0 0
ˆ ˆ
2 2
y x
ev I ev I
F i j
x x
 
 
   

0 0
2 2
y y
x
x x
e Iv e Iv
F
F a
x m mx
 
 
       
or 2 2
0
0
.
2
x
x x
dv e I
v v v
dx mx


  
0
2 2
0
2
x x
x
v dv e I dx
m x
v v



 

at maximum separation, max , x
x x v O
 
from eqn. (i) y O
v v

Now integrating
0
max
0
0
2
0
0
2 2
0
2
mv
x
Ie
x x
v a
x
v dv e I dx
x ae
m x
v v





  

 

8. Application : A particle with specific charge
q
m
moves in the region of
space where there are uniformly mutually perpendicular
electric and magnetic fields with strength ˆ
Ej and
induction ˆ
Bk . At moment 0
t  the particle was located
at the point O and had zero velocity. Find
y
x
z
E
B O
(i) The law of motion ( )
x t and ( )
y t of the particle, the shape of the trajectory.
(ii) The length of the segment of the trajectory between two nearest point at which
the velocity of the particles turns into zero.
Solution : Let the velocity of charge particle at P is
ˆ ˆ
x y
V v i v j
 

Lorentz force on the charged particle at point
P is ˆ
ˆ ˆ ˆ
( )
x y
F qE qV B qEj q v i v j Bk
      
   
ˆ ˆ ˆ
x y
F qEj qv Bj qv Bi
   

and x
x y
qE qv B
qB
a a
m m

   …(i)
y vy v
vx
P
x
differentiating eqn. (i)
y x
x y
da dv
qB qB qB qB
a v
dt m dt m m m
 
       
 
and
2 2
2
y
y
da q B
v
dt m
  or
2 2
2 2
y
y
d v qB
v
dt m
  …(ii)
Solution of above equation (ii)
sin
y
qB
v A t
m

 
 
 
 
…(iii)
where A and  are constant
According the problem, at 0,
t  0, , 0, 0
x y x y
qE
a a v v
m
   
sin( ) 0
o A o
  
    
sin
y
v A t

 
Differentiating, cos cos
y
y
dv
A t a A t
dt
   
  
at 0
t  ,
qE qE
A A
m m


  
Substituting value of A and  in equation …(iii)
sin
y y
qE qE
v t v
qB
m m
m


  
sin
y
E
v t
B

  …(iv)
Now,
0
(1 cos )
t
y
E
y v dt t
B


  
 …(v)
x component , sin
x y
qE qB E
a v t
m m B

   
     
   

9. sin
x
qE
a t
m

 
Integrating
0 0
sin (1 cos )
x
v t
x x
qE qE
dv t v t
m m
 

   
 
or (1 cos ) (1 cos )
x x
qE E
v t v t
qB B
m
m
 
    
 
 
 
Now,
0
0
sin
. [ sin ]
t
t
x
E t E
x v dt t t t
B B

 
  
 
    
 
 

2
( sin )
mE
x t t
qB
 
  
Shape of trajectory :
here [ sin ] sin
E B x
x t t t t
B E

   

    
and [1 cos ] cos 1
E B y
y t t
B E

 

    
we know 2 2
sin cos 1
t t
 
 
2 2
1 1
B x B y
t
E E
 

   
    
   
   
Solving we will get
2 2 2
2 2
E E E
x t y
B B B
 
   
   
   
    x
y
O
This is an equation of circle of radius
E
R
B
 , its center moves with constant velocity 0
E
v
B
 .
The path of particle is cycloid shown in figure, defined circumference of circle of radius
E
R
B
 .
(ii) Instantaneous speed of charged particle is
2 2
x y
V v v
 
Putting the value of x
v and y
v we will get
2
sin
2
E t
v
B


velocity becomes zero after on time period
2 m
T
qB


so,
2 /
2
0 0
2 2 8
sin sin .
2 2
m qB
T
E t E t mE
S dt S
B B qB

 
   
 

10. F
FO
OR
RC
CE
E O
ON
N A
A C
CU
UR
RR
RE
EN
NT
T C
CA
AR
RR
RY
YI
IN
NG
G C
CO
ON
ND
DU
UC
CT
TO
OR
R :
:
1. Force on current element idl


( )
dF i dl B
 

 
 
Proof d
V 

drift velocity
Force on 1
1 , ( )( )
d
e dF e V B
  

  
n  no. of e per unit volume
dl
B
i
e
dl
B
vd
i
B
B
2. Total force on e in volume ( , ), ( )( )( )
A dl dF e V B Adl n
  
 
( ) ( )
d
dF eAnv dl B dF i dl B
     

 
 
 

 
direction of magnetic force on determined by
(a) Right hand palm rule
(b) Flemmings left hand rule
3. Total force on current carrying conductor
(i) In uniform magnetic field B

 
( )
F i dl B i dl B
   
 

 

  
( )
F i l B
  

 
(ii) In non-uniform magnetic field
F dl B
 


 
B
dl
B
i
l
B
Application : A straight wire of mass 200 g and length 1.5 carries a current of 2A. It is suspended in
mid-air by a uniform horizontal magnetic field B. What is the magnitude of the magnetic
field ?
i
mg
l
B
Solution : We have that of mid-air suspension
mg ilB

mg
B
il

0.2 9.81
0.65
2 1.5
T

 

.
Application : Calculate the force on a current carrying conductor in a uniform magnetic field as shown.

11. × × × × × × × × × × × × × × × × × × × × × × × × × × ×
× × × × × × × × × × × × × × × × × × × × × × × × × × ×
× × × × × × × × × × × × × × × × × × × × × × × × × × ×
× × × × × × × × × × × × × × × × × × × × × × × × × × ×
× × × × × × × × × × × × × × × × × × × × × × × × × × ×
× × × × × × × × × × × × × × × × × × × × × × × × × × ×
× × × × × × × × × × × × × × × × × × × × × × × × × × ×
× × × × × × × × × × × × × × × × × × × × × × × × × × ×
× × × × × × × × × × × × × × × × × × × × × × × × × × ×
× × × × × × × × × × × × × × × × × × × × × × × × × × ×
× × × × × × × × × × × × × × × × × × × × × × × × × × ×
B
A
a
b
I
Q
I
c
P
R
S
r
T
d
I B
P
Solution: The net force from A to B. ( )
dF I dL B
 
  
Q
B P R T B
1 2 3 4 5
A A P Q R T
dF I[dL B] I[dL B] I[dL B] I dL B I(dL B)
 
         
 
     
         
The entire path can be broken down into elemental vectors joined to each other in
sequence. We know, from polygon law of addition of vectors what vector joining the tail
of the first vector to the head of the last vector is the resultant.
( )
F I L B
 
  
; where 2 2
| | 2
L a c b r d
    

2 2
. ( 2
net
F I B a c b r d
      and direction is upwards in the plane of paper.
AMPERE’S LAW:
Similar to the Gauss’s law of electrostatics, this law provides us short cut methods of finding
magnetic field in cases of symmetry. According to this law, the line integral of magnetic field
over a closed path ( . )
B dl



is equal to 0
 times the net current crossing the area enclosed by that
path. Mathematically, 0
( . )
net
B dl i





Positive direction of current and the direction of the line integral are given by the right hand
thumb and curling fingers respectively.
In order to find magnetic field using Ampere’s law the closed path of lone integral is generally
chosen such the B

is either || or  to the path line. Also, wherever B

is || to the path, its value
should be constant.
Brain Teaser: A long, straight wire carries a current. Is Ampere’s law valid for a loop that does not
enclose the wire? That encloses the wire but is not circular?
Brain Teaser: In order to have a current in a long wire, it should be connected to a battery or some such
device. Can we obtain the magnetic field due to a straight, long wire by using Ampere’s
law without mentioning this other part of the circuit?
Application: Suppose that the current density in a wire of radius a varies with r
according to , where K is a constant and r is the distance from the
axis of the wire. Find the magnetic field at a point distance r from
the axis when (I) r a and (II) r < a.
Solution: Choose a circular path centered on the conductor’s axis and apply
Ampere’s law.
(i) To find the current passing through the area enclosed by
the path

12. 2
( )(2 )
IdA Kr rdr

 i.e.,
4
3
0
2
2
t
kr
I k r dr


 

Since
4
0 0
, 2 .
2
kr
B dl I B r

  
  



3
0
4
kr
B

 
(ii) If r > a, then net current through the Amperian loop is
a 4
2
0
Ka
I Kr 2 rdr
2

   

Therefore B =
4
0
4
Ka
r

.
THE AMPERE:
Two current carrying straight conductors placed near each other will exert (magnetic) forces on
each other due to magnetic field of each other.
Figure shows two long parallel conductors separated by a distance d and bearing currents a
i and
b
i . The conductor ‘a’ produces a magnetic field a
B at all points along the conductor ‘b’. The
right-hand rule (figure) tells us that the direction of this field is downwards. its magnitude is given
by or form Ampere’s circuital law,
d
2
a
0



i
Ba
a
Fba
Ba
b
d
L
ib
ia
Two long straight parallel conductors carrying steady currents a
i and b
i and separated by a
distance d. a
B is the magnetic field set up by conductor ‘ a’ at conductor ‘b’.
The conductor ‘b’ will experience a sideways force on account of the external field a
B . The
direction of this force is towards the conductor ‘a’. You can verify this either by the cross product
rule of vectors or by Fleming’s left hand rule which is depicted in figure. We label this force ba
F ,
the force on a segment L of ‘b’ due to ‘a’. The magnitude of this force is given by,
a
b
ba LB
i
F 
d
2
i
i b
a
0




13. It is of course possible to compute the force on ‘a’ due to ‘b’. From considerations similar to
above we find that the force ba
F , on a segment of length L of ‘a’ due to the current in ‘b’ is equal
in magnitude to ba
F and directed towards ‘b’. Thus,
ab
ba F
F 

Note :
Parallel currents attract, and anti-parallel currents repel:
Let ba
B represent the magnitude of the force ba
F per unit length. Then, form,
0
2
a b
ba
i i
F
d



The ampere is the value of that steady current which, when maintained in each of the two very
long, straight, parallel conductors of negligible cross-section, and placed one meter apart in
vacuum, would produce on each of these conductors a force equal to 2 × 10-7 newtons per metre
of length.
When a steady current of 1A is set up in a conductor, the quantity of charge that flows through its
cross-section in 1s is one coulomb (1C).
Brain Teaser: Two wires carrying equal currents i each, are placed perpendicular to each other, just
avoiding a contact. If one wire is held fixed and the other is free to move under magnetic
forces, what kind of motion will result?
Brain Teaser: Two current-carrying wires may attract each other. In absence of other forces, the wires
will move towards each other increasing the kinetic energy. Does it contradict the fact
that the magnetic force cannot do any work and hence cannot increase the kinetic energy?
Application: The horizontal component of the earth’s magnetic field at a certain place is 3.0 × 10-5 T
and the direction of the field is from the geographic south to the geographic north. A very
long straight conductor is carrying a steady current of 1A. What is the force per unit
length on it when it is placed on a horizontal table and the direction of the current is
(a) east to west;
(b) south to north?
Solution: F I B
l
 

 
F I Bsin
l
 
(a) When the current is flowing from east to west,
0
90


Hence
5
(1 )(1 ) (3 10 )
F IlB A m T

  
The direction of the force is downwards. This direction may be obtained either by
Fleming’s left hand rule or the directional property of cross product of vectors.
(b) When the current is flowing from south to north,
0
0


F = 0
Hence no force per unit length on the conductor.
TORQUE ON A CURRENT CARRYING PLANAR LOOP IN A UNIFORM MAGNETIC FIELD

14. Case 1:
When plane of the loop is perpendicular to magnetic field
Length of AB = DC = 
And that of BC = AC = b
Force experienced by all the sides are shown in the figure
 Force on AB and DC are equal and opposite to the each other and that on BC and AD too.
0
F
  
Since the line of action of the forces on AB and DC is same and also the line of action of the
forces BC and AD is same, therefore torque is zero.
A B
C
D
BIb
BIb
BIl
BIl

Case 2:
When the plane of the loop is inclined to the magnetic field.
In this case again 0
F
 
A
 Lines of action of the forces on AB and DC are different, therefore this forms a couple
and produces a torque. Side view of the loop is shown in the figure.
B


b sin 
F = BIl
F = BIl
Torque = BI l (b sin  ) = BI(lb) sin 
 BIA sin .
If loop has N turns then


 sin
BNIA
In vector form
,
B where NIA
  
  


  
(is known as magnetic moment of the loop) A

is the area vector of the loop whose magnitude is
area of the loop and direction is out of the plane for anti-clock wise sense of the current and into
the loop for clockwise work of current.

15. Energy needed to rotate the loop through an angle d is dU = d
 


















2
1
2
1
d
sin
B
d
dU
U
1 2
(cos cos ),
U B
  
   if we choose at 1
 such that at 1 1
, 0
U
 
 
This is the energy stored in the loop.
.
m
U B

 


.
Brain Teaser: A rectangular current loop is in arbitrary orientation in an external magnetic field. Is any
work required to rotate the loop about an axis perpendicular to its plane?
Application: The rectangular coil having 100 turns is turned in a uniform
magnetic field of
0.05 ˆ
2
j . Tesla as shown in the figure. Find
the torque acting on the loop.
Solution: The magnetic dipole moment
of the current carrying coil is
given by ˆ
m NIAn


z
x
y
-0.04m
0.08m
I=0.5A
ˆ
100 0.5 0.08 0.04i
   
2 2
1.6 10 Am

 
The torque acting on the coil is
m B
  



ˆ ˆ
mB(i j)
 
2 0.05 ˆ
16 10 k
2

  
4 ˆ
5.66 10 (N m)k

  