Lecture 38 Intor_Biot Savart Law _CH4_L-1_18.07.2023.pptx

Purvesh kumar|IRPS jagdishpur
purveshkumar@gmail.com LECTURE 38
CLASS XII
MOVING CHARGES & MAGNETISM CH-4 L-1
Chapter 4
Moving Charges & Magnetism
Oersted’s Experiment
Magnetic Field
BIOT SAVART Law

CLASS XII
Oersted’s Experiment
Hans Christian Oersted (1777-1851) showed that electricity and
magnetism are related to each other.
His research later used in radio, television etc.
OERSTED EXPERIMENT: Oersted performed an experiment in 1820
that laid the very basis of relationship between electric current and magnetic
field.
He connected a simple circuit containing a cell, battery and key (switch). A
magnetic compass was placed near the circuit as shown in the figure.

CLASS XII
He observed that as soon as the key of the circuit gets closed the needle of
magnetic compass deflected.
He then reversed the direction of the current in the circuit and closed the
key, again the magnetic needle deflected but this time in opposite direction.
Oersted showed that electricity and magnetism are closely related
phenomena and concluded that electric current can produce magnetic field.
The direction of that magnetic field depends on the direction of the electric
current.

CLASS XII
MAGNETIC FIELD: It is the region surrounding a magnet, in which the
magnetic force of attraction or repulsion can be detected.
Strength of magnetic field is denoted by B.
It is a vector quantity, having both direction & magnitude.
SI unit of magnetic field is Tesla (T)
CGS unit is Gauss 1 T = 10–4 G
MAGNETIC FIELD LINES: These are the imaginary lines which gives
the direction and strength of the magnetic field.
The tangent to the magnetic field line at a point gives the direction of
magnetic field at that point.

CLASS XII
Right Hand Thumb Rule or Curl Rule: I
B
If a current carrying conductor is imagined to be
held in the right hand such that the thumb points
in the direction of the current, then the tips of
the fingers encircling the conductor will give the
direction of the magnetic lines of force.
This is known as Right Hand Thumb Rule.
It is used to find the direction of magnetic
field by a current carrying conductor

Magnetic Field is coming out of the plane of the paper.
Magnetic Field is going inside the plane of the paper.

CLASS XII
BIOT SAVART LAW
x
The strength of small magnetic field dB due to a
small current element dl carrying a current I at a point
P lying at a distance r from the current element dl is
directly proportional to I,
directly proportional to current element dl,
directly proportional to sin of angle between current
element dl and vector r, &
inversely proportional to the square of the distance
(r2) of point P from the current element dl.
θ
P
dl
r
P’
I

CLASS XII
where θ is the angle
between dl and r.
(ii) dB dl

(i) dB I

(iii) dB sinθ

2
1
(iv) dB
r

Mathematically According to Biot Savart Law:
Combining Eqn (i), (ii), (iii) & (iv)
2
I dl sinθ
dB
r

0
2
μ I dl sinθ
dB
4π r

Where μ0 is constant of proportionality
known as Permeability of the Vacuum
Value of μ0
x
θ
P
dl
r
P’
I
7 1 1 1
0
μ 4π 10 Tm A Wbm A
or
   
 

CLASS XII
Biot – Savart’s Law in vector form:
Direction of dB is perpendicular to plane containing dl and r
0
2
μ I dl sinθ
ˆ
dB n
4π r

0
3
μ I dl r sinθ
ˆ
dB n
4π r

0
3
μ I dl r
dB
4π r


Vector Form of Biot-Savart
Law
Where is a unit vector
perpendicular to plane
containing dl and r
n̂
Current element is a vector quantity whose magnitude is the vector product
of current and length of small element having the direction of the flow of
current. ( I dl)

CLASS XII
Magnetic Field Due to a Current
Carrying Straight Wire
+
Numerical Problems

CLASS XII
where θ is the angle
between dl and r.
(ii) dB dl

(i) dB I

(iii) dB sinθ

2
1
(iv) dB
r

Mathematically According to Biot Savart Law:
Combining Eqn (i), (ii), (iii) & (iv)
2
I dl sinθ
dB
r

0
2
μ I dl sinθ
dB
4π r

Where μ0 is constant of proportionality
known as Permeability of the Vacuum
Value of μ0
x
θ
P
dl
r
P’
I
7 1 1 1
0
μ 4π 10 Tm A Wbm A
or
   
 
RECAP

CLASS XII
Magnetic field due to an infinitely long current carrying straight conductor
Consider an infinitely long straight conductor carrying a
current I as shown in figure.
P be a point at a distance ‘a’ from the conductor.
AB is a small current element of length dl at a distance l
from the mid point O of the wire.
θ is the angle between the current element Idl and the line
joining the element dl and the point P.
r be the distance of point P from the length element dl.
According to Biot- Savart law, the small magnetic field dB
at the point P due to the current element Idl is
0
2
μ I dl sinθ
dB .....(1)
4π r

Not In Syllabus

CLASS XII
0
2
μ I dl sinθ
dB .....(1)
4π r

2 2
.....(2)
r l a
 
From Δ OAP
Also
sinθ sin(π θ)
a
r
   2 2
.....(3)
a
l a


Using (2) & (3) in equation (1)
0
2 2 3/2
μ Idl
dB
4π ( )
a
l a


Total magnetic field B due to the straight conductor
is integration of small magnetic field dB.
B dB


 
0
2 2 3/2
μ Idl
4π ( )
a
l a




 0
2 2 3/2
0
μ I dl
2
4π ( )
a
l a

 
  

 

Not In Syllabus

CLASS XII
0
2 2 3/2
0
μ I dl
2π ( )
a
l a

 
  

 

Let tanθ
l a

2
dl sec θ dθ
a
 
Equation becomes
2
/2
0
2 2 2 3/2
0
μ I sec θ dθ
2π ( tan θ )
a a
B
a a

 
  

 

2
/2
0
2 2 3/2
0
μ I sec θ dθ
2π [ (1 tan θ)]
a a
a

 
  

 

2
/2
0
2 2 3/2
0
μ I sec θ dθ
2π [ sec θ]
a a
a

 
  
 

2
/2
0
3 3
0
μ I sec θ dθ
2π sec θ
a a
a

 
  
 
  
/2
0
2 0
μ I
cosθ dθ
2π
a
a

 
1
θ tan
l
a
  
  
 
When 0; θ 0
When ; θ /2
l
l 
 
  
Not In Syllabus

CLASS XII
 
/2
0
2 0
μ I
cosθ dθ
2π
a
a

 
 
/2
0
0
μ I
sinθ
2πa


0
μ I
sin sin 0
2π 2
a

 
 
 
 
 
0
μ I
B 1 0
2πa
 
0
μ I
B
2πa

Expression of
Magnetic field at a
distance a for a long
straight wire
0
μ I
B
4πa

Expression of
Magnetic field at a
distance a for a semi
infinite straight wire
Not In Syllabus
Expressions are in Syllabus

CLASS XII
SOLVED EXAMPLES:
Example 1:
A wire placed along the N-S direction carries a current of 8 A from south
to north. Find the magnitude of magnetic field due to a 1 cm piece of wire
at a point 200 cm north-east from the piece.
Solution: The figure can be drawn as:
0
2
μ I dl sinθ
dB
4π r

By Biot Savart Law
Here I= 8A; dl = 1cm = 10-2m;
r = 200cm = 2m; θ = 45°
7 -2
2
4π 10 8 10 sin45
dB
4π 2

  

7 -2
10 8 10 0.707
4

  
 7 -2
10 2 10 0.707

   
9
dB 1.414 10 T

 

CLASS XII
Example 2:
A long straight wire carrying a current of 30A is placed in an external uniform
magnetic field of induction 4 × 10−4T parallel to the direction of the current. Find
the magnitude of the resultant magnetic field at a point 2 cm away from the wire.
Solution: The figure can be drawn as:
0
1
μ I
B
2πa

Magnetic field due to wire is:
Here I= 30A; a = 2cm = 2×10-2 m;
7
-2
4π 10 30
2π 2 10

 

 
4
1
B 3 10 T

 
2cm
Both B1 and B are perpendicular to each other
Net Magnetic field Bnet is given as:
2 2
net 1
B B +B

4 2 4 2
(3 10 ) +(4 10 )
 
  
8
25 10
  4
net
B 5 10 T

 

CLASS XII
THANK YOU
Feel free to mail me your queries at
purveshkumar@gmail.com

Lecture 38 Intor_Biot Savart Law _CH4_L-1_18.07.2023.pptx

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