## 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).

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.

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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