Lorentz Force Magnetic Force on a moving charge in uniform Electric and Magnetic fields

1. Priyanka Jakhar
Physics Lecturer
GGIC Vijay Nagar
Ghaziabad U.P.
Class 12th
Physics
MAGNETIC EFFECT OF CURRENT
Lorentz Force
Magnetic Force on a moving charge in uniform
Electric and Magnetic fields

2. MAGNETIC EFFECT OF CURRENT - II
1. Lorentz Magnetic Force
2. Fleming’s Left Hand Rule
3. Force on a moving charge in uniform Electric and Magnetic fields
4. Force on a current carrying conductor in a uniform Magnetic Field
5. Force between two infinitely long parallel current-carrying
conductors

3. Magnetic force :--- force on a moving charge in Magnetic field .
When a charged particle moves in a Magnetic field ,a force acts on it which is known as
magnetic force.
Let a test charge q is projected in a magnetic field B from a point in different direction with
different velocities, Tt was observed that magnitude of magnetic force (F )acting on the test
charge
1) is directly proportional to magnitude of charge .
2) is directly proportional to magnetic field strength .
3) directly proportional to the component of velocity perpendicular to magnetic field .
If theta is the angle between velocity of charge ( v ) and magnetic field ( B ) .
F ∝ 𝒗 𝑰
F ∝ 𝒒
F ∝ 𝑩
Or
F ∝ 𝒗𝒔𝒊𝒏𝜽

4. From equation 1 , 2 and 3
F ∝ 𝒒 𝒗 𝑩𝒔𝒊𝒏𝜽
F = 𝒌 𝒒 𝒗 𝑩𝒔𝒊𝒏𝜽
Where K is constant experimentally the value of k is 1 in SI system .
Special Cases:
i) If the charge is at rest, i.e. v = 0, then F = 0.So, a stationary charge in a magnetic field does
not experience any force.
ii) If θ = 0° or 180° i.e. if the charge moves parallel or anti-parallel to the direction of the magnetic
field, then
F = 0
iii) If θ = 90° i.e. if the charge moves perpendicular to the magnetic field, then the force is
maximum.
m
F = q v Bm (max)
m
F = 𝒒 𝒗 𝑩𝒔𝒊𝒏𝜽

5. Lorentz Magnetic Force:
A current carrying conductor placed in a magnetic field experiences a force . Which means that a
moving charge in a magnetic field experiences force.
F = q (v x B)
m
+q B
v
F
I
θ -
q
B
v
F
θF = (q v B sin θ) n
m
where θ is the angle between v and B
or
I
Value of magnetic force is zero in following conditions
1) If q = 0 ,the particle is neutral .
2) If v = 0 ,the particle is stationary.
3) If B = 0 ,magnetic field is zero .
The direction of magnetic force acting on a charge charged particle is perpendicular to its
velocity . A constant and uniform magnetic field can neither accelerate nor retard the charged
particle. It can only change the direction of the motion of the particle. Hence in a constant and
uniform magnetic field the kinetic energy of a moving charge particle remains same. In other
words no work is done by the magnetic force on the charged particle.

6. The direction of magnetic force acting on a charged particle depends on the sign also.
If q is positive charge magnetic force will be along the direction of
And if q is negative charge magnetic force will be opposite to
To find the direction of force acting on the moving positively charged
particles can be determined by one of the following is laws
( v x B )
( v x B )
Right hand screw rule :--if vector and are the places that their
tails are inside if a screw is placed perpendicular to and and is rotated
from ,the direction of motion of tip of Screw is the direction of
or direction of force 𝑭 on positive charge.
𝒗 𝑩
𝑩𝒗
𝒗 𝒕𝒐 𝑩 𝒗 × 𝑩 +
I
θ
𝒗
𝑩
𝑭
Right hand rule :-- vector and are places that their tails are
coinciding . Now keeping the palm of right and perpendicular to and
thumb should be in the plane of palm and fingers are
perpendicular to it. Now fingers are folded from to .
In this condition the direction of thumb represents
of force on the positive charge.
𝒗
𝒗
𝒗
𝑩
𝑩
𝑩
𝒗 × 𝑩
𝑭
+
I
θ
𝒗
𝑩
𝑭

7. Fleming’s Left Hand Rule: Force
(F)
Magnetic
Field
(B)
Electric
Current
(I)
If the central finger, fore finger and thumb of left hand are stretched
mutually perpendicular to each other and the central finger points to
current, fore finger points to magnetic field, then thumb points in the
direction of motion (force) on the current carrying conductor.
TIP:
Remember the phrase ‘e m f’ to represent electric current, magnetic field and force in
anticlockwise direction of the fingers of left hand.
Remember the phrase ‘Father , Mother , Child ’ to represent force , magnetic field and current, in
clockwise direction of the fingers of left hand.
Definition , unit and dimension of :---To define at any point in magnetic field ,at test
charge q is projected in different directions with different velocity from that point and
measure the force acting on it
F = q v Bm (max)
F / q v = Bm (max)
𝑩 𝑩

8. Unit :--In SI system unit of force is Newton ,unit of charge is coulomb or ampere –second and
unit of velocity is metre/second.
Unit of B = =
Newton
( ampere –second ) ×( metre/second )
Newton
( ampere – metre )
This is also known as Tesla. Which is named after scientist Nikola Tesla .
From the definition of magnetic flux and other SI unit of B is Weber / metre .
1 tesla :--1 tesla is the magnitude of that magnetic field in which a charged particle of one coulomb moving
with velocity 1 metre per second perpendicular to field experiences a force of 1 Newton.
1 tesla = 1 newton / ampere-meter = 1weber / meter 2
2
Dimension :--- 𝑩 =
[ 𝑭 ]
𝒒 × [ 𝒗 ]
=
[ 𝑴 𝑳 𝑻−𝟐 ]
𝑨 𝑻 [ 𝑳 𝑻−𝟏 ]
= [ 𝑴 𝑻−𝟐 𝑨−𝟏]
Tesla is a big unit in practice, a smaller unit of magnetic field is gauss .

9. Force on a current carrying conductor in a magnetic field :--
When a current carrying conductor is placed in magnetic field, a magnetic force acts on the conductor .
let a b is a brass rod which is free to slide on two parallel brass rails M and N .
Friction between rod and the rails is zero.
Rails are placed in magnetic field which is perpendicular to the plane of the paper directed downward .
When no current is flowing through rod, the rod is stationary but when key K is closed and current I is passed
through the rod from A to B ,the rod starts sliding to towards right on a rails .
If the direction of current is reversed by interchanging the positive and negative terminal of the battery ,the rod
starts flight towards left .
It is clear that in magnetic field a force F acts on a current carrying conductor which is perpendicular to both
current and magnetic field B.
If magnetic field is established in plane of paper parallel to rails .Rod remains stationary no magnetic force acts
on the rod .
Ampere had conclude on the basis of different experiments that force on current carrying conductor in magnetic
field is
1) directly proportional to the current in the conductor .
2) directly proportional to the length of the conductor.
3) directly proportional to the magnetic field strength .
𝑭 ∝ 𝑩
𝑭 ∝ 𝑳
𝑭 ∝ 𝑰
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x x
xx
x
x x
x
x
F
E
K
a
b
I
M
N

10. 4) directly to the Sine of angle between length of conductor and magnetic field .
From 1, 2 , 3 and 4
Where K is proportional to be constant in SI system
k =1 .
This is the expression for force on a current carrying conductor placed in magnetic field.
In vector form
Where 𝑳 is a vector and it is directed along the direction of current .
Equation one and two are valid for straight current carrying conductor.
𝑭 is directed along which can be determined by the laws of vector multiplication.
If a conductor is placed perpendicular to magnetic field ,to find the direction of magnetic force
we can find them by following two Laws.
1) Fleming left hand rule
2) Right hand Palm rule number 2
𝑭 ∝ 𝒔𝒊𝒏𝜽
𝑭 ∝ 𝑰𝑩𝑳𝒔𝒊𝒏𝜽
𝑭 = 𝒌 𝑰𝑩𝑳𝒔𝒊𝒏𝜽
𝑭 = 𝑰𝑩𝑳𝒔𝒊𝒏𝜽 −− −𝟏
𝑭 = 𝑰 [𝑳 × 𝑩 ] ---------2
[𝑳 × 𝑩 ]

11. Current flows in a conductor due to motion of charged particles .
If conductor is situated in magnetic field on every charged particle a Lorentz magnetic force
will act.
Force on the conductor will be the vector sum of the magnetic force acting on the charged
particle.
Let a conducting metallic wire of length L and area of cross section A situated in uniform
magnetic field B length of conductor makes an angle 𝜽 with the direction of magnetic field .
As per free electron model, current will be due to motion of free charge in the metallic
conductor.
Let 𝒗 𝒅 is the Drift velocity of the electrons and e is the charge of electron .
Magnetic force on each electron
Magnitude of force
Force on a current carrying conductor on the basis of Force on a Moving charge:
𝑭 𝟏 = 𝒆𝒗 𝒅 𝑩𝒔𝒊𝒏𝜽
𝑭 𝟏 = −𝒆 [𝒗 𝒅 × 𝑩 ]

12. If n be the number density of electrons or n is the number of free electron per unit
volume.
A be the area of cross section of the conductor
Then no . of electrons in the element dl is n A dl.
Then no . of electrons in the conductor L is n A L .
In vector form
Drift velocity of each electron is same, magnetic force on each electron will be same.
total magnetic force acting on the conductor
F = n A L 𝑭 𝟏 F = n A L (𝒆𝒗 𝒅 𝑩𝒔𝒊𝒏𝜽 )
Where n𝒆𝑨𝒗 𝒅 = IF = I L 𝑩𝒔𝒊𝒏𝜽 −− −𝟏
𝑭 = 𝑰 [𝑳 × 𝑩 ] ------2
L
I
𝑩𝜽
e
e
ee
e
e
Important conclusions
i) If θ = 0° or 180° i.e. if the charge moves parallel or anti-parallel to the direction of the magnetic
field, then force is minimum
F = 0
ii) If θ = 90° i.e. if the charge moves perpendicular to the magnetic field, then the force is
maximum.
F = IB Lm (max)

13. θdl
F
I
I
B
A 𝑳 𝟏
-
iii) Equation one and two are valid for straight current carrying conductor.
iv) If a current carrying conductor of any shape or arbitrary shape is placed in a uniform
magnetic field.
Then the magnetic force on it will be calculated by dividing in into strips of small length and
we integrate them.
in this condition magnetic force acting on thin strip d 𝑳
force acting on the conductor
B is uniform
vi ) For a current carrying loop of arbitrary arbitrary shape
the magnetic force acting on a current carrying loop placed in a uniform magnetic field is
zero.
d 𝑭 = 𝑰 [d 𝑳 × 𝑩 ]
d 𝑭 = 𝑰 [ d 𝑳 ] × 𝑩
d 𝑳 = 𝑳 𝟏
𝑭 = 𝑰 [ 𝑳 𝟏 × 𝑩 ]
𝑩
d 𝑳 = 𝟎

14. Unit :--In SI system unit of force is Newton ,unit of current is ampere and unit of velocity is
metre.
Unit of B =
Newton
( ampere ) ×( metre )
This is also known as Tesla. Which is named after scientist Nikola Tesla .
From the definition of magnetic flux and other SI unit of B is Weber / metre .
1 tesla :--1 tesla is the magnitude of that magnetic field in which on keeping a conductor of 1 meter length
having 1 ampere current in it experiences a force 1 Newton .
1 tesla = 1 newton / ampere-meter = 1weber / meter 2
2
Dimension :--- 𝑩 =
[ 𝑭 ]
𝑰 × [ 𝑳 ]
=
[ 𝑴 𝑳 𝑻−𝟐 ]
𝑨 [ 𝑳]
= [ 𝑴 𝑻−𝟐 𝑨−𝟏]
Tesla is a big unit in practice, a smaller unit of magnetic field is gauss .
B = F / I Lm (max)

15. Motion of a charged particle in a uniform magnetic field
Let a positively charged particle of charge q and enters in uniform magnetic field 𝑩 with velocity 𝒗 at an angle θ
The force acting on a particle
Condition 1
If θ = 0° or 180° i.e. when velocity of particle is parallel or anti-parallel to the direction of the magnetic
field, then force is minimum
𝒔𝒊𝒏𝜽 = 0 F = 0 path of the particle will be straight line.
θ = 0° θ = 180°
Condition 2
If θ = 90° i.e. when velocity of particle is perpendicular to the magnetic field, then the force is
maximum.
𝒔𝒊𝒏𝜽 = 1
At every point of a path, magnitude of force will be constant and direction of the force will be perpendicular to
the velocity. This is the condition for circular motion.
Hence the path of the particle moving perpendicularly to the magnetic field will be circular.
The necessary centripetal force required to move on a circular path will be provided by a magnetic force .
𝒗 × 𝑩 F = 𝒒 𝒗 𝑩𝒔𝒊𝒏𝜽𝑭 = q [ ]
F = 𝒒 𝒗 𝑩 = Fm (max)
𝑩
𝑩
𝒗 𝒗

16. If m is the mass of the particle and r is the radius of the circle path
𝒎𝒗 𝟐
𝒓
= F = q v B
r =
𝒎𝒗
𝒒𝑩
If magnetic field is directed perpendicular to plane of the paper in downward direction and
positively charged particle is moving in a plane of a paper at different points of the magnetic
force will be directed as .
The particle moves in a circular path in anticlockwise direction.
If P is the momentum of particle ,
P = m v
radius circular path of a charged particle is directly proportional to momentum of the particle.
if K is the kinetic energy of the particle
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
xx
x
x
x
x
x xx
x
x x
x
x
x
x
x
x
x x
x
x
x x
x
x
x
x
x
𝑭
𝑭
𝑭
𝑭
𝒗
𝒗
𝒗
𝒗
K =
𝟏
𝟐
m𝒗 𝟐 =
𝒑 𝟐
𝟐𝒎
r =
𝟐𝒎 𝒌
𝒒𝑩
P = 𝟐𝒎 𝒌

17. Direction of magnetic field
Perpendicular to plane of paper in
downward direction.
Perpendicular to plane of paper in
downward direction .
Perpendicular to plane of paper in
upward direction .
Perpendicular to plane of paper in
upward direction .
Moving charge (in
the plane of paper)
Positive +
Negative -
Positive +
Negative -
Circular path
Anticlockwise
Clockwise
Clockwise
Anticlockwise
x
x
.
.

18. r =
𝟐𝒎 𝒌
𝒒𝑩
K = q V if particle is accelerated by potential V volt
r =
𝟐𝒎q V
𝒒𝑩
Time period
time period of circular path
substituting the value of r
Frequency
Angular frequency
T =
𝟐𝝅𝒓
𝒗
T =
𝟐𝝅
𝒗
𝒎𝒗
𝒒𝑩 T =
𝟐𝝅𝒎
𝒒𝑩
𝒇 =
𝟏
𝑻
=
𝒒𝑩
𝟐𝝅𝒎
𝝎 = 𝟐 𝝅 𝒇 = 𝟐 𝝅 𝒒𝑩
𝟐 𝝅 𝒎
=
𝒒𝑩
𝒎

19. Condition 3
when angle between velocity of particle and magnetic field θ is other then 0°,180°and 90° .
In this condition velocity we can be resolved into two mutually perpendicular components 1)
1) Component parallel to magnetic field B , 𝒗 𝑰𝑰 = 𝒗 𝒄𝒐𝒔𝜽
2) Component perpendicular to magnetic field B , 𝒗⊥ = 𝒗 𝒔𝒊𝒏𝜽
particle will move on a straight line path due to a component parallel to magnetic field.
where is due to perpendicular component it will move in a circular path .
The resultant of these two is found in the form of helix whose axis is parallel to magnetic field.
In The Helix path of the circle will be perpendicular to the field.
Radius of the helix r =
𝒎𝒗⊥
𝒒𝑩
=
𝒎𝒗 𝒔𝒊𝒏𝜽
𝒒𝑩
Time period
Pitch :--- distance covered by the particle along the direction of field in completing one
circular path is known as pitch of the helix it is represented by 𝒑.
T =
𝟐𝝅𝒎
𝒒𝑩
𝒑 = 𝒗 𝑰𝑰 × 𝑻 = 𝒗 𝒄𝒐𝒔𝜽 ×
𝟐𝝅𝒎
𝒒𝑩
=
𝟐𝝅𝒎𝒗 𝒄𝒐𝒔𝜽
𝒒𝑩

20. Force on a moving charge in uniform Electric and Magnetic Fields:
When a charge q moves with velocity v in region in which both electric field E and magnetic field B
exist, then the Lorentz force is
F = qE + q (v x B)
or F = q (E + v x B)
When a charge q remains is stationary in an electric field E then force on it due to electric field will be F = q E .
When a charge q moves with velocity in a magnetic field by making an angle 𝜽 with the direction of
then on it due to field is given by
𝒗 𝑩 𝑩
F = 𝒒 𝒗 𝑩𝒔𝒊𝒏𝜽
𝒗 × 𝑩𝑭 =
𝑭 𝑬 + 𝑭 𝑩𝑭 =
F = 𝒒𝑬 + 𝒒 𝒗 𝑩𝒔𝒊𝒏𝜽
If the particle enters in a region where both electric field and magnetic field are present then it experiences force
due to both fields (which are independent of each other) and the total force acting on the charged particle is
called Lorentz Force.

21. 1) The term which is contributed by the electrical is known as Lorentz electric force. F = q E
The term contributed by the magnetic field is known as Lorentz magnetic force .
2) The Lorentz electric force is independent of velocity of charge while Lorentz magnetic force depends upon its
velocity .
3) When all the three are collinear then only will produce acceleration in the motion of charge while is
given by 𝒂 =
𝑭 𝑬
𝒎
=
𝒒𝑬
𝒎
The direction of this acceleration will be in the direction of electric field so speed ,velocity , Momentum and
kinetic energy all will change .
4) When all mutually perpendicular are and magnitude and direction of 𝑬 and 𝑩 are adjusted in such a
manner that magnitude of electric and magnetic force become equal and opposite then net force on the
particle will be zero and particle will pass through the field undeviated .
or q E = 𝒒 𝒗 𝑩𝒔𝒊𝒏𝜽 if 𝜽 = 90˚
This condition is used to select the particles moving with certain velocity from the beam of moving charges
moving with different velocity .
so mutually perpendicular electric and magnetic field work as a velocity selector .
In this condition only the particles with velocity E / B will pass through crossed electric and magnetic field
undeflected.
𝒗 × 𝑩𝑭 =
𝒗 ,𝑩 , E
𝒗 ,𝑩 , 𝑬
or 0 = qE + q (v x B)𝑭 𝑬 + 𝑭 𝑩 = 0𝑭 =
𝑭 𝑬 = 𝑭 𝑩
v =
𝑬
𝑩

22. Forces between two parallel infinitely long current-carrying conductors:
Magnetic force between two parallel current carrying conductor can be calculated using
following relationship
electric current magnetic field electric current
Every current carrying conductor produce magnetic field around itself .
A magnetic force acts on a current carrying conductor placed in magnetic field .
Let two parallel wires 1 and 2 are placed in the air at a distance of d from each other in the
plane of paper 𝑰 𝟏 and 𝑰 𝟐 are the current following flowing in the wires in same direction
therefor magnetic field at a point on the wire 2 due to wire 1.
from Maxwell right hand screw rule 𝑩 𝟏 will be directed perpendicular to the plane of the
paper in downward direction .
Since magnetic field on this wire due to wire 2 will be zero .Therefore on every point of wire 2
the net magnetic field will be 𝑩 𝟏 . Wire 2 having current 𝑰 𝟐 is placed in magnetic field
𝑩 𝟏 perpendicularly .
Therefore force acting on length L of the wire 2
𝑩 𝟏 =
𝝁 𝟎 𝑰 𝟏
𝟐𝝅𝒅

23. r
F21F12
I1
P
Q
I2
S
R
B =1
μ I0 1
2π r
Magnetic Field on RS due to current in PQ is
Force acting on RS due to current I through it is2
F21 =
μ I0 1
2π r
I l sin 90˚2
B acts perpendicular and into the plane of the diagram by Right
Hand Thumb Rule. So, the angle between l and B is 90˚ . l is
length of the conductor.
1
1
F21 =
μ I I l0 1 2
2π r
B =2
μ I0 2
2π r
Magnetic Field on PQ due to current in RS is
Force acting on PQ due to current I through it is1
F12 =
μ I0 2
2π r
I l sin 90˚1
F12 =
μ I I l0 1 2
2π r
(The angle between l and
B is 90˚ and B Isemerging out)2 2
F12 = F21 = F =
μ I I l0 1 2
2π r
F / l =
μ I I
2π r
0 1 2
or
or
Force per unit length of the conductor is
N / m
(in magnitude)
(in magnitude)
x B 1
B2

24. r
F
I1
P
Q
F
I2
x
S
R
r I2
F
x
S
R
I1
F
P
Q
x
By Fleming’s Left Hand Rule,
the conductors experience
force towards each other and
hence attract each other.
By Fleming’s Left Hand Rule,
the conductors experience
force away from each other
and hence repel each other.
Case-2: Parallel wire carrying currents in the opposite direction:
•On proceeding in the similar manner as the first case, we will find that the values of forces will be
the same, only their directions get reversed (refer the diagram above)
•The forces will be equal but this time away from each other, i.e., the wires will move away from
each other (repel each other)

25. Definition of Ampere:
F / l =
μ I I
2π r
0 1 2
Force per unit length of the conductor
is
N / m
When I = I = 1 Ampere and r = 1 m, then F = 2 x 10 N/m. One ampere is that
current which, if passed in each of two parallel conductors of infinite length
and placed 1 m apart in vacuum causes each conductor to experience a force
of 2 x 10 Newton per metre of length of the conductor.
1 2
-7
-7
Representation of Field due to Parallel Currents:
I1 I2
B
I1 I2
B
N