# Magnetic field due to straight conductor carrying current

Consider a straight conductor carrying current ‘i’. Let ‘P’ be a point at a perpendicular distance ‘a ’ from the conductor. Let ‘dl’ be a small current element at a distance ‘r’ from ‘P’.

According to Biot-Savart’s law, the magnetic induction at P due to the small element is

$\large dB = \frac{\mu_0}{4\pi} \frac{I dl sin\phi}{r^2}$   …(i)

In ΔPOC , θ + φ = 90°  ⇒ φ = 90- θ

sinφ = sin(90- θ) = cosθ

$\large cos\theta = \frac{a}{r}$

$\large r = a sec\theta$

$\large tan\theta = \frac{l}{a}$

$\large l = a tan\theta$

On differentiating ,

$\large dl = a sec^2\theta d\theta$

From (i)

$\large dB = \frac{\mu_0}{4\pi} \frac{I (a sec^2\theta d\theta) cos\theta}{a^2 sec^2 \theta}$

$\large dB = \frac{\mu_0}{4\pi} \frac{I}{a}cos\theta d\theta$

On integrating ,

$\large B = \frac{\mu_0}{4\pi} \frac{I}{a} \int_{-\theta_1}^{\theta_2} cos\theta d\theta$

$\large B = \frac{\mu_0}{4\pi} \frac{I}{a} (sin\theta_1 + sin\theta_2)$

Special cases :

Case I: If the wire extends to infinity on either side of ‘ o ‘ then

θ1 = θ2 = π/2

$\displaystyle B = \frac{\mu_0}{4 \pi} \frac{I }{R} ( sin\frac{\pi}{2} + sin\frac{\pi}{2})$

$\displaystyle B = \frac{\mu_0}{4 \pi} \frac{2 I}{R}$

Case II: If length of the wire is finite say ‘ L ‘ and P lies on right bisector of wire, then

$\displaystyle \theta_1 = \theta_2 = \theta = sin^{-1}(\frac{L}{\sqrt{4R^2 + L^2}} )$

$\displaystyle B = \frac{\mu_0}{4 \pi} \frac{ I}{R} (2 sin\theta)$

In this way we can find magnetic field at any point due to straight current.

Example : Two semi-infinitely long straight current carrying conductors are in form of an ‘ L ‘ shape as shown in the figure. The common end is at the origin. What is the value of magnetic field at a point (a, b), if both the conductors carry the same current I?

Solution : For the conductor along the X- axis, the magnetic field

$\displaystyle B_1 = \frac{\mu_0}{4 \pi} \frac{I }{b} ( sin\theta_2 + sin\frac{\pi}{2} )$ ; along the negative Z-axis

$\displaystyle B_1 = \frac{\mu_0}{4 \pi} \frac{I }{b} ( \frac{a}{\sqrt{a^2 + b^2}} + 1 )$

For the conductor along Y-axis, the magnetic field is

$\displaystyle B_2 = \frac{\mu_0}{4 \pi} \frac{I }{a} ( sin\theta_1 + sin\frac{\pi}{2} )$ ; along the negative Z-axis

$\displaystyle B_2 = \frac{\mu_0}{4 \pi} \frac{I }{a} ( \frac{b}{\sqrt{a^2 + b^2}} + 1 )$

The net magnetic field is,

$\displaystyle \vec{B} = \vec{B_1} + \vec{B_2}$

$\displaystyle B = \frac{\mu_0}{4 \pi} \frac{I }{b} ( \frac{a}{\sqrt{a^2 + b^2}} + 1 ) + \frac{\mu_0}{4 \pi} \frac{I }{a} ( \frac{b}{\sqrt{a^2 + b^2}} + 1 )$

$\displaystyle B_2 = \frac{\mu_0 I}{4 \pi a b} ( a + b + \sqrt{a^2 + b^2} )$

Example : A current I is established in a closed loop of an triangle ABC of side l . Find the magnetic field at the centroid ‘ O ‘

Solution:

$\displaystyle OP = \frac{l}{2\sqrt3}$

Magnetic fields due to current in all three sides are equal in magnitude and directed into the plane of the paper.Hence net field ,

$\displaystyle B = 3 \frac{\mu_0 I}{4 \pi r} ( sin\frac{\pi}{3} + sin\frac{\pi}{3} )$

Where, $\displaystyle r = \frac{l}{2\sqrt3}$

$\displaystyle B = 3 \frac{\mu_0 I}{4 \pi r} ( 2 sin\frac{\pi}{3} )$

$\displaystyle B = 9 \frac{\mu_0 I}{4 \pi l}$

Exercise : In the figure shown two infinitely long parallel straight current carrying wires are separated by a distance d. The current in each wire is I. Points A and B are in the plane of the wires. Find the ratio of the magnetic field BA at A and BB at B.