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.
