Notes-Part-2-Class-12-Physics-Chapter-10-Magnetic Fields due to Electric Current-MSBSHSE

Expression for the magnetic induction near a straight infinitely long current-carrying wire :

Suppose a point P is at a distance a from a straight, infinitely long, wire carrying a current I, as shown in below Fig.

The incremental magnetic induction \(\vec{dB}\) at the point P due to a current element, \(I\vec{dl}\)  is

  \(\vec{dB}\) = \((\frac{μ_0}{4π})\frac{I\vec{dl}×\hat{r}}{r^2}\)….(1)

In magnitude,

dB = \((\frac{μ_0}{4π})\frac{Idlsinθ}{r^2}\)    ….(2)

At the point P, \(\vec{dB}\)  is directed perpendicular to the plane of the figure and into of the page as given by the right hand rule for the direction of \(\vec{dl}×\hat{r}\)

At point P, \(\vec{dB}\) has this same direction for all the current elements into which the wire can be divided. Thus, we can find the magnitude of the magnetic field produced at P by the current elements in the lower half of the infinitely long wire by integrating dB in Eq. (2), from 0 to ∞.

Now consider a current element in the upper half of the wire, one that is as far above P as \(I\vec{dl}\) is below P. By symmetry, the magnetic field produced at P by this current element has the same magnitude and direction as that from \(I\vec{dl}\)  in Fig. Thus, the magnetic field produced by the upper half of the wire is exactly the same as that produced by the lower half. Hence, the magnitude of the total magnetic field at P is

B  = \(\int_{-∞}^{∞}dB\) = \(2\int_{0}^{∞}dB\)

     = \(2\frac{μ_0I}{4π}\int_{0}^{∞}\frac{dlsinθ}{r^2}\)     …..(3)

From fig r = \(\sqrt{l^2+a^2}\)

∴ sin θ = \(\frac ar \) = \(\frac{a}{\sqrt{l^2+a^2}}\)

∴ B = \(2\frac{μ_0I}{4π}\int_{0}^{∞}\frac{dl}{(l^2+a^2)}\frac{a}{\sqrt{l^2+a^2}}\)

   = \(\frac{μ_0aI}{2π}\int_{0}^{∞}\frac{dl}{(l^2+a^2)^{3/2}}\)    …..(4)

   = \(2\frac{μ_0aI}{4π}×\frac{1}{a^2}[\frac{l}{(l^2+a^2)^{1/2}}]_{i=0}^{i=∞}\) 

    = \(\frac{μ_0}{4π}\frac{2I}{a}[1-0]\)=\(\frac{μ_0}{4π}\frac{2I}{a}\)  …..(5)

That is, the magnitude B is inversely proportional to the distance from the wire. Because of the axial symmetry about the straight wire, the magnetic induction has the same magnitude B at all points on a circle in a transverse plane and centred on the conductor; the direction of \(\vec{B}\) is everywhere tangential to such a circle. Thus, the magnetic field lines around the current in the straight wire is a family of circles centred on the wire.

Expression for the force per unit length between two infinitely long parallel conductors carrying current and hence define the ampere :

When two currents pass in adjacent parallel straight conductors, we may think of each of the currents as being situated in the magnetic field caused by the other current. This results in a force on each conductor.

Consider two infinitely long, straight parallel wires, each of length l a distance s apart in vacuum, as shown in Fig. (a). The magnetic field around the wire 1, carrying a current I₁ has an induction of magnitude

B₁ = \((\frac{μ_0}{4π})\frac{2I_1}{s}\)

Wire 2, with a current I₂ in the same direction as I₁, is situated in this field. The direction of the field with induction \(\vec{B_1}\) at the position of wire 2, given by the right hand [grip] rule, is perpendicular to the plane of the two conductors, as shown. Hence, the force \(\vec{F_2}\) on wire 2 has a magnitude.

F₂ = I₂ l B₁ = \((\frac{μ_0}{4π})\frac{2I_1I_2l}{s}\)   …..(1)

and is, by Fleming's left hand rule, towards wire 1. Similarly, the magnetic induction \(\vec{B_2}\)  at the position of wire 1 has a magnitude

B₂ = \((\frac{μ_0}{4π})\frac{2I_2}{s}\)

and is also directed perpendicular to the plane of the wires. Hence, the force   on wire 1 has a magnitude

F₁ = I₁lB₂ = \((\frac{μ_0}{4π})\frac{2I_1I_2l}{s}\)  …..(2)

directed towards wire 2. Thus, the two currents attract each other. \(\vec{F_1}\) = \(-\vec{F_2}\)  i.e., they are equal in magnitude and opposite in direction.

Ampere found that the wires attracted each other when the currents in them were in the same direction [Fig. (a)], and repelled each other when they were in the opposite directions [Fig. (b)].

From the Eq. (2), the force per unit length acting on each wire is \(\frac{F}{l}\) = \((\frac{μ_0}{4π})\frac{2I_1I_2}{s}\)

Using SI units, μ₀/4π = 10⁻⁷ N/A² and, if I₁ =I₂ = 1 A and s = 1m, then

\(\frac{F}{l}\) = 2 x 10⁻⁷  N/m

In SI, this equation is the defining relation for the ampere.

Magnetic Field Produced by a Current in a Circular Arc of a Wire:

Below figure depicts a circular arc of a wire (AB), carrying a current I.

First obtain the magnetic field produced by one current-length element of the arc and then integrate over the entire arc length.

The circular arc AB subtends an angle θ at the centre O of the circle of which the arc is a part, and r is its radius.

The unit vector \(\hat{r}\) from each current element \(I\vec{dl}\) towards the centre of the loop is perpendicular to \(I\vec{dl}\), i.e., the angle between them is 90°.

Since every current element is equidistant from the centre of curvature O, the magnetic field at O due to each current element in the arc by Biot-Savart’s law is

\(\vec{dB}\) = \((\frac{μ_0}{4π})\frac{I\vec{dl}×\hat{r}}{r^2}\) ……(1)

dB = \((\frac{μ_0}{4π})\frac{Idl\,sin90^0}{r^2}\)= \((\frac{μ_0}{4π})\frac{Idl}{r^2}\)   ……(2)

Equation (2) gives the magnitude of the field.

The direction of the field is given by the right hand rule. Aligning the thumb in the direction of the current, the field direction is indicated by the curling fingers. Thus, the direction of each of the dB is into the plane of the paper. The total field at O is therefore,

B = ∫dB = \(\frac{μ_0}{4π}I\int_{A}^{B}\frac{dl}{r^2}\)

   = \(\frac{μ_0}{4π}I\int_{0}^{θ}\frac{r}{r^2}dθ\)

   =  \(\frac{μ_0}{4π}\frac{Iθ}{r}\)…..(3)

where the angle θ is in radians.

An electric current in a circular loop establishes a magnetic field similar in every respect to the field of a magnetic dipole (or a bar magnet).

Consider a circular conducting loop of radius R, axis along the x-axis and carrying a current I. The area of the loop is A = πR² and \(\vec{A}\) has the direction given by right hand rule. The axial magnetic induction of the current loop at a distance x from its centre is

\(\vec{B}\) = \((\frac{μ_0}{4π})\frac{2I\vec{A}}{(R^2+z^2)^{3/2}}\) = \((\frac{μ_0}{4π})\frac{2\vec{μ}}{(R^2+z^2)^{3/2}}\)

where \(\vec{M}\)= \(I\vec{A}\)  is the magnetic moment of the current loop and μ₀ is the permeability of free space.

For x >> R, ignoring R² in comparison with x²,

\(\vec{B}\) = \((\frac{μ_0}{4π})\frac{2\vec{μ}}{x^3}\)

This equation also gives the magnetic induction on the axis of a short magnetic dipole (or a bar magnet) of magnetic moment \(\vec{μ}\)

For a magnetic dipole, the dipole moment is directed from the south pole of the dipole to its north pole. For a current loop, the magnetic dipole moment has the direction of the axial field of the current loop as given by the right-hand rule.

When an observer looking at a current carrying circular loop finds the direction of the current anticlockwise, the face of the loop towards the observer acts as the north-pole.

When an observer looking at a current-carrying circular loop finds the direction of the current clockwise, the face of the loop towards the observer acts as the south- pole. This rule is known as the clock rule.

Explanation : Figure shows two wires carrying currents I₁ and I₂ in vacuum. The magnetic induction \(\vec{B}\) at any point is the net effect of these currents.

To find the magnitude B of the magnetic induction:

• We construct an imaginary closed curve around the conductors, called an Ampérian loop, and imagine it divided into small elements of length dl. The direction of \(\vec{dl}\) is the direction along which the loop is traced.
• We assign signs to the currents using the right hand rule : If the fingers of the right hand are curled in the direction in which the loop is traced, then a current in the direction of the outstretched thumb is taken to be positive while a current in the opposite direction is taken to be negative.

For each length element of the Ampérian loop, \(\vec{B}.\vec{dl}\) gives the product of the length dl of the element and the component of \(\vec{B}\) parallel to \(\vec{dl}\). If θ is the angle between \(\vec{dl}\) and \(\vec{B}\),

\(\vec{B}.\vec{dl}\) = (B cos θ) dl

Then, the line integral, \(\oint \vec{B}.\vec{dl}\) = \(\oint B\, cos θ\, dl\)

For the case shown in above Fig. the net current I through the surface bounded by the loop is

I = I₂ − I₁

∴ \(\oint B\,cos θ\, dl\) = μ₀I  = μ₀ (I₂ − I₁)   …….(3)

Equation (3) can be solved only when B is uniform and hence can be taken out of the integral.

Remeber : Ampére's law in magnetostatics plays the part of Gauss's law of electrostatics. In particular, for currents with appropriate symmetry, Ampére’s law in integral form offers an efficient way of calculating the magnetic field. Like Gauss’s law, Ampere’s law is always true (for steady currents), but it is useful only when the symmetry of the problem enables B to be taken out of the integral \(\oint B\, cos θ\, dl\). The current configurations that can be handled by Ampére’s law are infinite straight conductor, infinite plane, infinite solenoid and toroid.

Expression for the magnetic induction near a current-carrying straight, infinitely long wire.

Consider a point P at a distance a from a straight, infinitely long wire carrying a current I in free space, see Fig. (a).

Because of the axial symmetry about the straight wire, the magnetic induction has the same magnitude B at all points on a circle in a transverse plane and centred on the wire. We, therefore, choose an Ampérian loop a circle of radius a centred on the wire with its plane perpendicular to the wire, as shown in above Fig. (b).

\(\vec{B}\) is everywhere tangential to the circular Ampérian loop. Therefore, the angle θ between \(\vec{B}\) and a length element \(\vec{dl}\) is zero at all points of the loop.

∴ \(\oint \vec{B}.\vec{dl}\) = \(\oint B\,dl\,cos θ\) = \(B\oint dl\) = B(2πa)

since cos θ = 1 and B has the same value around the path. \(\oint dl\)  gives the circumference of the circular loop.

In the figure, the Amperian loop is traced in the anticlockwise sense, so that the current I is taken as positive in accordance with the right hand rule.

By Ampére’s law,

\(\oint \vec{B}.\vec{dl}\) = μ₀I

∴ B(2πa) = μ₀I

∴ B = \(\frac{μ_0I}{2πa}\)

This is the required expression. .

Expression for the magnetic induction inside an ideal solenoid carrying a steady current :

An ideal solenoid is tightly wound and infinitely long. Let n be the number of turns of wire per unit length and I be the steady current in the solenoid.

For an ideal solenoid, the magnetic induction  inside is reasonably uniform over the cross section and parallel to the axis throughout the volume enclosed by the solenoid; \(\vec{B}\) outside is negligible.

As an Ampérian loop, we choose a rectangular path abcd of length l parallel to the solenoid axis, Fig. The width of the rectangle is taken to be sufficiently large so that the side dc is far from the solenoid where \(\vec{B}\) = 0. The line integral of the magnetic induction around the Amperian loop in the sense abcda is

\(\oint \vec{B}.\vec{dl}\) = \(\oint_{a}^{b} \vec{B}.\vec{dl} + \oint_{b}^{c} \vec{B}.\vec{dl} + \oint_{c}^{d} \vec{B}.\vec{dl} + \oint_{d}^{a} \vec{B}.\vec{dl}\) ……(1)

\(\vec{B}\) is constant inside and is parallel to side ab.

Hence, as we go in the same direction as \(\vec{B}\)  from a to b, \(\vec{B}\) and \(\vec{dl}\) are parallel so that

\(\oint_{a}^{b} \vec{B}.\vec{dl}\) = \(\oint_{a}^{b} Bdl\) = \(B\oint_{a}^{b} dl\) = \(Bl \) ……(2)

Along the paths c → d and d → a, \(\vec{B}\) is perpendicular to \(\vec{dl}\) inside the solenoid while \(\vec{B}\) = 0 outside.

∴ \(\oint_{b}^{c} \vec{B}.\vec{dl}\) = \(\oint_{d}^{a} \vec{B}.\vec{dl}\) = 0 ……(3)

Also \(\vec{B}\) = 0 along side cd, so that

\(\oint_{c}^{d} \vec{B}.\vec{dl}\) = 0  ……(4)

Thus, from Eqs. (1), (2), (3) and (4),

\(\oint \vec{B}.\vec{dl}\) = \(Bl \)  ……(5)

The total current enclosed by the Amperian loop is

I_end = current through each turn x number of turns enclosed by the loop

      = I x nl = nl I   ……(6)

By Ampere’s law,

\(\oint \vec{B}.\vec{dl}\)   = μ₀ I_end       (in vacuum)

Therefore, from Eq.s (5) and (6),

Bl = μ₀ nl I

∴ B = μ₀ nI    ….(7)

This is the required expression.

Remember :

• The field inside an ideal solenoid is uniform— it doesn't depend on the distance from the axis. In this sense, the solenoid is to magnetostatics what the parallel-plate capacitor is to electrostatics; a simple devicefor producing strong uniform fields.
• At an axial point at the end of a long solenoid, B = ½ μ₀ nI   
• Although the above result for B is obtained for an ideal solenoid, it is also valid for a realistic solenoid, particularly when applied to points in the middle of it but certainly not to points near the ends. Thus, a solenoid can be designed for a specific value of B by a choice of i and n.

Expression for the magnetic induction inside an ideal toroid carrying a steady current :

An ideal toroid consists of a long conducting wire wound tightly around a torus made of a nonconducting material. When a steady current is passed through it, the magnetic induction  in the interior of the toroid is tangent to any circle concentric with the axis of the toroid and has the same value on this circle.

Suppose the toroid has N turns of wire and I is the current in its coil. As our Ampérian loop, we choose a circle of radius r concentric with the axis of the toroid, as shown in Fig. Since \(\vec{B}\) has the same value on this circle and is tangential to it, we go around this path in the direction of \(\vec{B}\) so that \(\vec{B}\) and \(\vec{dl}\) are parallel. Then, the line integral of the magnetic induction around the Ampérian loop is

\(\oint \vec{B}.\vec{dl}\) = \(\oint B\,dl\) = \(B\oint dl\) = B(2πr)    …..(1)

The net current enclosed by the Amperean loop is

I_end = current through each turn x number of turns enclosed by the loop

      = I x N = NI   ……(2)

By Ampere’s law,

\(\oint \vec{B}.\vec{dl}\)  = μ₀ I_end      ..... (in free space)

Therefore, from Eq.s (1) and (2),

B(2πr) = μ₀ NI

∴ B = \(\frac{μ_0NI}{2πr}\)  

This is the required expression.