(ii)
Magnetic field due to a straight current carrying conductor of finite length
Suppose AB is a straight conductor carrying a current of I and magnetic field intensity is
to be determined at point P. According to Biot-Savart law the magnetic field at P
B
I
F
G
dl
90° d
 r 
Q
R
E


P
I
A
 
0 I dl  r

dB = 4
r3

angle between I dl and r is (180 – ) so
Now
dB =
0 I dl sin 180   
4
r2
dB =
 0 I dl sin 
4
r2
...(i)
EG = EF sin 
= dl sin 
and
EG = EP sin d  rsin d
= r d
so
dl sin  = r d
...(ii)
so from (i)
dB =
 0 Id
4 r
...(iii)
from  EQP,
so
R
r = cos 
dB =
 0 I cos  d
4
R
...(iv)
Then the total magnetic field at point P due to the entire conductor is
2

B=
1
=
B 
0 I
cos  d
4 R
0 I
sin 
4 R
2
1
0 I
 sin 1  sin 2 
4 R
...(v)
for any conductor of infinite length
1 =  2  90
so
B=
0 2I
4 R
B 
0 I
NA 1 m 1
2 R
The direction of magnetic field due to a current carrying conductor can be obtained by
using any of the laws like
(i)
Right hand palm rule no 1
(ii)
Right hand thumb rule or
(iii) Maxwell Right-hand screw Rule.