Nature of magnetic field of long straight conductor

Consider a straight conductor XY carrying current I in the direction shown in Fig. It is desired to find the magnetic field at point P located at a perpendicular distance a from the conductor (i.e. PQ = a).

 Magnetic Field Due to Straight Conductor Carrying Current

Consider a small current element of length dl. Let r→ be the position vector of point P from the current element and θ be the angle between dl and r (i.e., ∠POQ = θ). Let us further assume that           QO = l.
According to Biot-Savart law, the magnitude of magnetic field dB→ at point P due to the considered current element is given by ;
dB =(µo/4π)((I*dl*sinθ)/r^2)  ...(i)
To get the total magnetic field B, we must integrate eq. (i) over the whole conductor. As we move along the conductor, the quantities dl, θ and r change. The integration becomes much easier if we express everything in terms of angle f shown in Fig. In the right angled triangle PQO, θ=90°–Ф, sinθ = sin(90°–Ф) = cosФ

Also, cosФ =a/r or r=a/cosФ

Further, tanФ=l/a or l=a*tanФ or
 => dl = a*sec^2ФdФ.
Putting the values of sin θ, dl and r in eq. (i), we have,
   dB = (µo/4π)((I*(a*sec^2ФdФ)*cosФ)/(a/cosФ)^2)
=> dB = (µo/4π)((I*cosФ dФ)/a) ...(ii)

The direction of dB→ is perpendicular to the plane of the conductor and is directed inwards (Right-hand grip rule). Since each current element contributes to the  magnetic field in the same direction, the total magnetic field B at point P can be found by integrating eq. (ii) over the length XY i.e.
=> B =∫(-Ф1,Ф2)dB=(µo/4π)(I/a)*∫(-Ф1,Ф2)cosФdФ
=> B =(µo*I/4πa)(sinФ2+sinФ1)...(iii)

Also, H = B/µo = (I/4πa)(sinФ2+sinФ1)
Eq. (iii) gives the value of B at point P due to a conductor of finite length.

Special cases:-
(i) When the conductor XY is of infinite length and point P lies at the centre of the conductor. In this case, Ф1 = Ф2 = 90° = π/2.
B =  (µo*I/4πa)(sinФ2+sinФ1)
B = (µo*2I/4πa)
Also, H = (B/µo) = (1/4π)(2I/a)=I/2πa
(ii) When conductor XY is of infinite length but point P lies near one end Y (or X). In this case, Ф1 = 90° and Ф2 = 0°.
B = [(µo*I)/(4πa)] (sin90 + sin0 )
B = [(µo*I)/(4πa)]
Note that it is half of that for case (i).
Also,  H = (B/µo) = (1/4π)(2I/a)=I/2πa
(iii) If the length of the conductor is finite (say l) and point P lies on the right bisector of the conductor. In this case, Ф1 = Ф2 = Ф .
Now,
sinФ=l/(4(a^2)+(l^2))^(1/2)
B=(µo*I/4πa)(sinФ+sinФ)=(µo*2I/4πa)(sinФ)=(µo*2I/4πa)(l/(4(a^2)+(l^2))^(1/2))
Also,
H = (B/µo) = (I/2πa)*(l/(4(a^2)+(l^2))^(1/2))

Direction of B  . For a long straight conductor carrying current, the magnetic lines of force are concentric circles with conductor as the centre ; the direction of magnetic lines of force can be found by right-hand grip rule. The direction of B at any point is along the tangent to field line at that point as shown in Fig.
 long straight conductor carrying current

Note. For a given current, B ∝ 1/a so that graph between B and a is a hyperbola.
NOTE:- VECTORS ARE BOLD OR WITH → SYMBOL.
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Last modified: Sunday, 22 September 2019, 7:41 PM