**THE MAGNETIC FIELD:**

In earlier lessons we found it convenient to describe the interaction between charged objects in terms of electric fields. Recall that an electric field surrounding an electric charge. The region of space surrounding a moving charge includes a magnetic field in addition to the electric field. A magnetic field also surrounds a magnetic substance. In order to describe any type of field, we must define its magnitude, or strength, and its direction.

**Magnetic field :** Magnetic field is the region surrounding a moving charge in which its magnetic effects are perceptible on a moving charge (electric current). Magnetic field intensity is a vector quantity and also known as magnetic induction vector. It is represented by $\vec{B}$.

Lines of magnetic induction may be drawn in the same way as lines of electric field. The number of lines per unit area crossing a small area perpendicular to the direction of the induction being numerically equal to $\vec{B}$.

The number of lines of $\vec{B}$ crossing a given area is referred to as the magnetic flux linked with that area. For this reason $\vec{B}$ is also called magnetic flux density.

There are two methods of calculating magnetic field at some point. One is Biot-Savart law which gives the magnetic field due to an infinitesimally small current carrying wire at some point and the another is Ampere’ law, which is useful in calculating the

magnetic field of a symmetric configuration carrying a steady current.

The unit of magnetic field is weber/m^{2} and is known as tesla (T) in the SI system.

### Biot-Savart’s Law :

In order to understand Biot-Savart law, we need to understand the term current-element. Current element is the product of current and length of an infinitesimal segment of a current carrying wire. The current element is taken as a vector quantity. Its direction is the same as the direction of current. The magnetic field (any such field) is defined at a point.

In the figure shown, there is a segment of a current carrying wire and P is the point where the magnetic field is to be calculated. $I\vec{dl}$ is a current element and $\vec{r}$ is the position vector of the point ‘ P ‘ with respect to the current element $I\vec{dl}$.

According to Biot-Savart’s Law, magnetic field at point P due to the current element $I\vec{dl}$ is

(i)is directly proportional to the current (I) flowing through the element .

(ii)is directly proportional to the length of the element .

(iii)is directly proportional to the sine of the angle (θ) between length of the element and the line joining the element to the point P.

(iv)is inversely proportional to the square of the distance (r) of the point from the element .

$ \displaystyle dB = \frac{\mu_0}{4 \pi} \frac{I dl sin\theta }{r^2}$

$ \displaystyle \vec{dB} = \frac{\mu_0}{4 \pi} \frac{I (\vec{dl} \times \vec{r})}{r^3}$

Since the entire segment is made-up of infinite such current elements and direction of magnetic field due to each element of P is same, the magnetic field due to the entire wire segment can be found by integrating the magnetic field due to the current elements over entire length of the wire.

Hence ,

$ \displaystyle \vec{B} = \int\frac{\mu_0}{4 \pi} \frac{I (\vec{dl} \times \vec{r})}{r^3}$

Here μ_{o} is called permeability of free space.

In SI units the value of ( $ \large \; \frac{\mu_0}{4 \pi}= 10^{-7} \;tesla-meter/ampere $ ) .

The SI units of magnetic fields are Tesla, (weber/m^{2})

In the above expression limits of the integral depend on the shape and size of the current carrying wire.

### 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

θ_{1} = θ_{2} = 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 B_{A} at A and B_{B} at B.