Magnetic Field due to Electric Current

Magnetic Field due to Electric Current

The magnetic field due to an electric current flowing through a wire can be calculated using Biot-Savart's law and Ampere's circuital law. Ampere's circuital law is used only in case of symmetry.

BIOT-SAVART'S LAW

To get the magnetic field at a point using Biot-Savart's law, we need to understand the term current-element. Current element is the product of current and length of infinitesimal segment of current carrying wire. The current element is a vector quantity. Its direction is same as the direction of current.

In the figure shown, there is a segment of a current carrying wire and P is a point where magnetic field is to be calculated. i$d\vec{\ell }$represents the current element and $\vec{r}$, the position vector of the point 'P' with respect to the current element I$d\vec{\ell }$. According to Biot-Savart's law, magnetic field at a point 'P' due to the current element i$d\vec{\ell }$is given by the expression, $d\vec{B}=\frac{{\mu }_o}{4\pi }\frac{Id\vec{\ell }\times \vec{r}}{r^3}$

Since entire segment in made-up of infinite such current elements, the magnetic field due to the entire wire segment can be found by integrating the magnetic field due to the current elements. Hence, ${\vec{B}}_{(atpo\mathrm{int}P)}=\int \frac{{\mu }_o}{4\pi }\frac{I\vec{d}\ell \times \vec{r}}{r^3}$

Here $\mu$_o is a proportionality constant, known as the permeability of free space.

In SI units, the value of $\mu$_o/4$\pi$ is equal to 10^-7 Tesla-meter/ampere.

The SI units of magnetic fields $\vec{B}$ are Tesla, weber/m².

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

MAGNETIC FIELD DUE TO CURRENT IN A STRAIGHT WIRE

The magnetic field due to a wire segment carrying current i at P, when the wire segment subtends angles $\alpha$ and $\beta$ as shown, can be determined as follows:

$\text{The magnitude of B = }\frac{{\mu }_{o}i}{4\pi R}\left[\cos \alpha +\cos \beta \right]$

In vector form $text\vec{B}=\frac{{\mu }_{o}i}{4\pi R}\left[\cos \alpha +\cos \beta \right]\left(-\hat{k}\right)$

If the wire is infinitely long then

$\alpha$ = 0 and $\beta$ = 0, hence, B = $\frac{{\mu }_{o}i}{2\pi R}$

Illustration 1: A current of 1A is flowing in the sides of an equilateral triangle of side 4.5 ´ 10 ^–2m. Find the magnetic field at the centroid of the triangle.

Solution: Let be the current flowing in the sides of triangle. The magnetic field due to one side (say BC) of triangle at centroid O

$B_1=\frac{{\mu }_{0}i}{4\pi R}(\sin \alpha +\sin \beta )$

Here i = 1A

$\frac{{\mu }_0}{4\pi }={10}^{-7}weber/A-m$

$\alpha$ = $\beta$ = 60°, R = OD = $\frac{BD}{\tan {60}^{\circ }}$

$=\frac{a/2}{\sqrt{3}}=\frac{a}{2\sqrt{3}}=\frac{4.5\times {10}^{-2}}{2\sqrt{3}}$m

$\therefore B_1=\frac{{10}^{-7}\times 1}{\left(4.5\times {10}^{-2}/2\sqrt{3}\right)}(\sin {60}^{\circ }+\sin {60}^{\circ })$

$=\frac{{10}^{-7}\times 2\sqrt{3}}{4.5\times {10}^{-2}}\left(\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\right)$

$=\frac{6}{4.5}\times {{10}^{-5}}_{}weber/m^2$

$\therefore$ The magnetic field due to whole triangle at centroid O is

$B=3B_1=\frac{3\times 6}{4.5}\times {10}^{-5}=4\times {{10}^{-5}}_{}weber/m^2$

MAGNETIC FIELD (B) AT THE CENTRE OF THE CURRENT CARRYING LOOP

Magnetic field induction B = $\frac{{\mu }_{o}i}{2R}$

MAGNETIC FIELD (B) AT THE CENTRE OF THE CURRENT CARRYING ARC

We have to find magnetic field at point P due to arc which subtends an angle q at the centre (P).

At the centre magnetic field induction B = $\frac{{\mu }_{o}i\theta }{4\pi R}$

Note: If the current in a loop is clockwise, then the magnetic field is directed inward perpendicular to the plane of the loop.

If the current in a loop is anticlockwise, then the magnetic field is directed outward perpendicular to the plane of the loop.

MAGNETIC FIELD ON THE AXIS OF A CURRENT CARRYING LOOP

Let the radius of the loop be a and axial distance be z.

Note that from symmetry, the resultant field at P must be along z, therefore :

$B_z=B=\frac{{\mu }_{o}i{a}^2}{2{(a^2+z^2)}^{3/2}}$ wb/m²

(i) B varies non linearly as shown in the plot between z and B and is maximum when z = 0, i.e. the point at the centre of the coil and then

B = $\frac{{\mu }_{0}i}{2a}$, Which is same as magnetic field at the centre of a circular coil

(ii) If z>>R, then

B = $\frac{{\mu }_{0}i{R}^2}{2z^3}$ = $\frac{{\mu }_0}{2\pi }\frac{i\pi R^2}{z^3}$= $\frac{{\mu }_0}{4\pi }\frac{i.2\pi R^2}{z^3}$

but pR² = A = Area cross section of the coil

Then B = $\frac{{\mu }_0}{4\pi }\frac{2iA}{z^3}=\frac{{\mu }_{0}2M}{4\pi z^3}$

Or, $\vec{B}=\frac{{\mu }_0}{4\pi }\frac{2\vec{M}}{z^3}$

Where $\vec{M}=i\vec{A}$= magnetic dipole moment of the current loop. The direction of $\vec{M}$ is same as the direction of normal of the area of the loop. If the current is in clockwise sense then direction of magnetic dipole moment is along inward normal, otherwise it is along outward normal.

If the coil consists of N turns then magnetic dipole moment of the loop will be given by M = NiA

Illustration 2: In bhor model of hydrogen atom, the electron circulates around the nucleus in a path of radius 5.1 ´ 10 ^–11m at a frequency 'v' of 6.8 ´ 10¹⁵ rev/s.

(a) What is the value of B at the centre of the orbit?

(b) What is the equivalent dipole moment?

Solution: (a) We know that $i=\frac{ch\arg e}{time}=ch\arg e\times frequency$

Here e = 1.6 x 10 ^–19C and v = 6.8 x 10¹⁵

$\therefore$ i = (1.6 x 10 ^–19) x (6.8 x 10¹⁵)

= 1.1 x 10 ^{– 3 amp.}

The field at the centre of a circular loop of current i is given by

$B=\frac{{\mu }_0}{4\pi }\frac{2\pi iN}{r}={10}^{-7}\times \frac{2\pi \times (1.1\times {10}^{-3})\times 1}{5.1\times {10}^{-11}}$

$=13.6weber/m^2\approx 14weber/m^2$

(b) Equivalent dipole moment = iA = i x $\pi$r²

= (1.1 x 10 ^{– 3})$\pi$ (5.1 x 10 ^{– 11})²

= 90 x 10 ^{– 24} Am²