OERSTED’S EXPERIMENT

The first evidence of the relationship of magnetism to moving charges was discovered in 1819 by the Danish scientist Hans Christian  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 IdL⃗ . The force on another similar conductor can be expressed conveniently in terms of a magnetic field dB⃗  due to the first. The properties of this magnetic field are as follows:

1. The magnetic field grows weaker as we move farther  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 IdL⃗ .
2. The larger the electric current, the larger is the magnetic field. In particular, the magnitude of the magnetic field dB⃗  is proportional to the
current I.

3. The magnitude of the magnetic field dB⃗  is proportional to sinθ, where θ is the angle between the current element IdL⃗    and vector r⃗  that points from the current element to the point in space where dB⃗  is evaluated. The direction of dL⃗   is the same as the direction of current at that point.
4. The direction of the B⃗  is not radially away from its source as the gravitational field and the electric field are from their sources. In fact, the direction of dB⃗  is perpendicular to both IdL⃗  and the vector r⃗ .
These features of field dB⃗  can be written compactly as

dB⃗ =(μ04π)IdL⃗ ×r^r2=(μ04π)IdL⃗ ×r⃗ r3

Here, (μ0/4π) is a constant of proportionality and r^ is unit vector in the direction of r⃗ .
The constant μ0 is called the permeability of free space or the permeability constant. Its value is

μ0=4π×10−7Tm/A

The magnetic field B⃗  is also called magnetic induction, or flux density.
Its SI unit is tesla (T) which is equivalent to Wb/m².
The dimensions of B⃗  are [MT ^– 2A^– 1]
The magnitude of dB⃗  can be obtained by

dB=(μ04π)IdLsinθr2

1. Field due to Straight Current-Carrying Conductor

According to Biot-Savart law,

dB⃗ =(μ04π)IdL⃗ ×r⃗ r3
⸫   B⃗ =(μ04π)∫IdL⃗ ×r⃗ r3

Case (A) : The point P is along the length of the wire, as in Fig. (A). Then dL⃗  and r⃗  will be either parallel or antiparallel. That is, θ = 0 or π, so dL⃗ ×r⃗ =0, hence  B⃗ =0      

Case (B) : The point P is at a  distance d from the wire, as shown in Fig. (B). Every current element IdL⃗  of the wire contributes to B⃗  in the same direction (which is ⊗). Therefore,

B=(μ04π)∫IdLrsinθr3=(μ04π)∫BAIdysinθr2

But y=dtanϕ⇒dy=dsec2ϕdϕ,     and     r=dsecϕ, θ=(90∘−ϕ).

⸫  B=(μ04π)Id∫α−βcosϕdϕ
or  B=(μ04π)Id(sinα+sinβ)

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∝1d  [and not 1d2].

(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, α=β=(π/2). Hence,

B=(μ04π)Id[1+1]   or   B=(μ04π)2Id

(5)  If  the point is near one end of an infinitely long wire, α=(π/2) and β=0. Hence,

B=(μ04π)Id[1+0]   or   B=(μ04π)Id

(6)  If the wire is of finite length and the point is on its perpendicular bisector, α=β. Hence,

B=(μ04π)2Idsinα  with   sinα=L(L2+4d2)√  where L is length of the wire.

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

B=(μ04π)Idsinα  with  sinα=L(L2+d2)√

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=μ04π.2Id=10−7×2×102.5×10−2=8×10−5=80μT                       

Total field due to both the wires is

2×80μT=160μT

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, 

B⃗ =B⃗ P+B⃗ Q=(μ04π)IRO⊙+(μ04π)ISO⊙
so B⃗ =(μ04π)2Id⊙           [as RO = SO = d]

Substituting the given data,
B⃗ =10−7×2×100.02⊙=10−4T⊙

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 dL⃗  and r⃗  is always π/2, and
(iii) the contribution of each element to B⃗  is in the same direction (i.e., out of the page if the current is anticlockwise and into the page if clockwise).

⸫   B⃗ =μ04π∫IdL⃗ ×r⃗ r3=μ04π∫BAIdLR2
But dL=Rdϕ,
⸫   B=μ04π∫α0IRdφR2       or  B=μ04πIαR

Note that
(1) The angle α is in radians
(2) If the loop is semicircular (i.e, α=π), B=μ04ππIR
(3) If the loop is a full circle with N turns (i.e., α=2πN), B=μ04π2πNIR