Chapter Notes: Magnetism & Moving Charge Physics Class 12

Notes for Magnetism and moving Charge chapter of class 12 physics. Dronstudy provides free comprehensive chapterwise class 12 physics notes with proper images & diagram.

OERSTED'S EXPERIMENT

The first evidence of the relationship of magnetism to moving charges was discovered in 1819 by the Danish scientist Hans ChristianÂ  Oersted. He found that a compass needle was deflected by a current carrying wire. When the direction of current was reversed the deflection also got reversed.

BIOT-SAVART'S LAW

It turns out that there is no source of magnetic force similar in nature to point electric charges giving rise to electric fields. Such magnetic monopoles have not been found so far. The elementary source of magnetic force is a current element $Id{\rm{\vec L}}$. The force on another similar conductor can be expressed conveniently in terms of a magnetic field $d{\rm{\vec B}}$Â due to the first. The properties of this magnetic field are as follows:

1. The magnetic field grows weaker as we move fartherÂ  from its source. In particular, the magnitude of the magnetic field dB is inversely proportional to the square of the distance from the current element $Id{\rm{\vec L}}$.
2. The larger the electric current, the larger is the magnetic field. In particular, the magnitude of the magnetic field $d{\rm{\vec B}}$Â is proportional to the
current I
.

3. The magnitude of the magnetic field $d{\rm{\vec B}}$ is proportional to sin$\theta$, where $\theta$ is the angle between the current element $Id{\rm{\vec L}}$Â  Â and vector $\vec r$Â that points from the current element to the point in space where $d{\rm{\vec B}}$Â is evaluated. The direction of $d{\rm{\vec L}}$Â Â is the same as the direction of current at that point.
4. The direction of the $\vec B$Â is not radially away from its source as the gravitational field and the electric field are from their sources. In fact, the direction of $d{\rm{\vec B}}$Â is perpendicular to both $Id{\rm{\vec L}}$Â and the vector $\vec r$.
These features of field $d{\rm{\vec B}}$Â can be written compactly as

$\,\,d\vec B = \left( {{{{\mu _0}} \over {4\pi }}} \right)I{{d\vec L \times \hat r} \over {{r^2}}} = \left( {{{{\mu _0}} \over {4\pi }}} \right)I{{d\vec L \times \vec r\,\,} \over {{r^3}}}$

Here, $\left( {{\mu _0}/4\pi } \right)$Â is a constant of proportionality and ${\rm{\hat r}}$Â is unit vector in the direction of $\vec r$.
The constant ${{\mu _0}}$Â is called the permeability of free space or the permeability constant. Its value is

${\mu _0} = 4\pi \times {10^{ - 7}}T{\rm{ }}m/A$

The magnetic field ${\rm{\vec B}}$Â is also called magnetic induction, or flux density.
Its SI unit is tesla (T) which is equivalent to Wb/m2.
The dimensions of ${\rm{\vec B}}$Â are [MT - 2A- 1]
The magnitude of $d{\rm{\vec B}}$Â can be obtained by

$\,\,dB = \left( {{{{\mu _0}} \over {4\pi }}} \right){{I{\rm{ }}dL\sin \theta } \over {{r^2}}}\,\,$

1. Field due to Straight Current-Carrying Conductor

According to Biot-Savart law,Â

$\,\,d\vec B = \left( {{{{\mu _0}} \over {4\pi }}} \right)I{{d\vec L \times \vec r\,\,} \over {{r^3}}}$
â¸« Â Â ${\rm{\vec B}} = \left( {{{{\mu _0}} \over {4\pi }}} \right)\int {{{I{\rm{ }}d{\rm{\vec L}} \times {\rm{\vec r}}} \over {{r^3}}}}$

Case (A) : The point P is along the length of the wire, as in Fig. (A). Then $d{\rm{\vec L}}$Â and ${\rm{\vec r}}$Â will be either parallel or antiparallel. That is, $\theta$ = 0 or $\pi$, so $d{\rm{\vec L}} \times {\rm{\vec r}} = 0$, hence Â ${\rm{\vec B}} = 0$Â Â Â Â Â Â

Case (B) : The point P is at aÂ  distance d from the wire, as shown in Fig. (B). Every current element $Id{\rm{\vec L}}$Â of the wire contributes to ${\rm{\vec B}}$Â in the same direction (which is $\otimes$). Therefore,

$B = \left( {{{{\mu _0}} \over {4\pi }}} \right)\int {{{I{\rm{ }}dL{\rm{ }}r\sin \theta } \over {{r^3}}}} = \left( {{{{\mu _0}} \over {4\pi }}} \right)\int_{{\rm{ }}A}^{{\rm{ }}B} {{{I{\rm{ }}dy{\rm{ }}\sin \theta } \over {{r^2}}}}$

But $y = dtan\phi \Rightarrow dy = dse{c^2}\phi d\phi$,Â Â Â Â  andÂ Â Â Â  $r = dsec\phi$, $\theta = (90^\circ - \phi )$.

â¸« Â $B = \left( {{{{\mu _0}} \over {4\pi }}} \right){I \over d}\int_{{\rm{ }} - \beta }^{{\rm{ }}\alpha } {\cos } \phi d\phi$
or Â $B = \left( {{{{\mu _0}} \over {4\pi }}} \right){I \over d}(\sin \alpha + \sin \beta )$

Note that
(1) For points along the length of the wire (but not on it) the field is always zero.
(2) For points at a perpendicular distance d from the wire, field B varies inversely with distance,
$B \propto {1 \over d}$ Â [and not ${1 \over {{d^2}}}$].

(3) The field is always perpendicular to the plane containing the wire and the point. So in a plane perpendicular to the wire the lines of force are concentric circles.
(4) If the wire is of infinite length and the point P is not near its any end, $\alpha = \beta = \left( {\pi /2} \right)$. Hence,

$B = \left( {{{{\mu _0}} \over {4\pi }}} \right){I \over d}[1 + 1]$ Â  or Â Â $B = \left( {{{{\mu _0}} \over {4\pi }}} \right){{2I} \over d}$

(5)Â  IfÂ  the point is near one end of an infinitely long wire, $\alpha = \left( {\pi /2} \right)$ and $\beta = 0$. Hence,

$B = \left( {{{{\mu _0}} \over {4\pi }}} \right){I \over d}[1 + 0]$ Â  or Â Â $B = \left( {{{{\mu _0}} \over {4\pi }}} \right){I \over d}$

(6)Â  If the wire is of finite length and the point is on its perpendicular bisector, $\alpha = \beta$. Hence,

$B = \left( {{{{\mu _0}} \over {4\pi }}} \right){{2I} \over d}\sin \alpha$ Â withÂ Â Â $\sin \alpha = {L \over {\sqrt {({L^2} + 4{d^2})} }}$ Â where L is length of the wire.

(7)Â  If the wire is of finite length and the point is near its one end, $\beta = 0$. Hence,

$B = \left( {{{{\mu _0}} \over {4\pi }}} \right){I \over d}\sin \alpha$ Â with Â $\sin \alpha = {L \over {\sqrt {({L^2} + {d^2})} }}$

Application 1
Figure shows two long straight wires carrying electric currents of 10 A each, in opposite directions. The separation between the wires is 5.0 cm. Find the magnetic field at a point P midway between the wires.

Solution:

The right-hand thumb rule shows that the magnetic field at P due to each of the wires is perpendicular to the plane of the diagram and is going into it. The magnitude of the field due to each wire is

$B = {{{\mu _0}} \over {4\pi }}{\rm{.}}{{2I} \over d} = {{{{10}^{ - 7}} \times 2 \times 10} \over {2.5 \times {{10}^{ - 2}}}} = 8 \times {10^{ - 5}} = 80\,\,\mu T$Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

Total field due to both the wires is

$2 \times 80\mu T = {\bf{160}}\mu {\bf{T}}$

Application 2
A pair of stationary and infinitely long bent wires are placed in the x-y plane as shown. The wires carry currents of 10 amperes each as shown. The segments L and M are along the x-axis. The segments P and Q are parallel to the y-axis such that OS = OR = 0.02 m. Find the magnitude and direction of the magnetic induction at the origin O.

Solution:

As point O is along the length of segments L and M so the field at O due to these segments will be zero. Further, as the point O is near one end of a long wire,Â

${\rm{\vec B}} = {{\rm{\vec B}}_P} + {{\rm{\vec B}}_Q} = \left( {{{{\mu _0}} \over {4\pi }}} \right){I \over {RO}} \odot + \left( {{{{\mu _0}} \over {4\pi }}} \right){I \over {SO}} \odot$
so ${\rm{\vec B}} = \left( {{{{\mu _0}} \over {4\pi }}} \right){{2I} \over d} \odot$Â Â Â Â Â  Â Â Â Â  [as RO = SO = d]

Substituting the given data,
${\rm{\vec B}} = {10^{ - 7}} \times {{2 \times 10} \over {0.02}} \odot = {10^{ - 4}}T \odot$

2. Field due to a Circular Current-Carrying Segment at its Centre

Let AB be a circular segment of radius R. Point P is at its centre. Here,
(i) each element is at the same distance from the centre, i.e., r = R = constant,
(ii) the angle between element $d{\rm{\vec L}}$Â and ${\rm{\vec r}}$Â is always $\pi /2$, and
(iii) the contribution of each element to ${\rm{\vec B}}$Â is in the same direction (i.e., out of the page if the current is anticlockwise and into the page if clockwise).

â¸« Â Â ${\rm{\vec B}} = {{{\mu _0}} \over {4\pi }}\int {{{I{\rm{ }}d{\rm{\vec L}} \times {\rm{\vec r}}} \over {{r^3}}}} = {{{\mu _0}} \over {4\pi }}\int_{{\rm{ }}A}^{{\rm{ }}B} {{{I{\rm{ }}dL} \over {{R^2}}}}$
But $dL = Rd\phi$,
â¸« Â Â $B = {{{\mu _0}} \over {4\pi }}\int_{{\rm{ }}0}^{{\rm{ }}\alpha } {} {{IR{\rm{ }}d\varphi } \over {{R^2}}}$Â Â Â Â Â Â  or Â $\,\,B = {{{\mu _0}} \over {4\pi }}{{I\alpha } \over R}\,\,$

Note that
(1) The angle $\alpha$ is in radians
(2) If the loop is semicircular (i.e, $\alpha = \pi$),Â $B = {{{\mu _0}} \over {4\pi }}{{\pi I} \over R}$
(3) If the loop is a full circle with N turns (i.e., $\alpha = 2\pi N$),Â $B = {{{\mu _0}} \over {4\pi }}{{2\pi NI} \over R}$

1 Comment

• Anonymous

Provide Full notes of this chapter