Ampere's Circuital Law

The line integral of the magnetic field around a closed loop is proportional to the current linked by the loop. The proportionality constant is equal to $\mu$_o (=4$\pi$ x 10^-7 H/m).

Mathematical form of Ampere's law is as follows

$\oint \vec{B}.d\vec{\ell }={\mu }_{0}i$

This law is useful for us only in the case of symmetry. For unsymmetrical cases the Biot-Savart law is more useful.

MAGNETIC FIELD (B) DUE TO AN INFINITELY LONG STRAIGHT CONDUCTOR

In the figure there is a long straight wire perpendicular to the plane of the paper. The inward direction of the current is shown by a cross. There is a point P at a distance r from the wire where the magnetic field is to be calculated. Draw a circle of radius r and symmetrically around the wire. This loop also represents magnetic lines of force due to the wire. The sense of lines of force can be found by using the right hand grip rule. At any point on this loop, $\overrightarrow{B}$ is tangential. d$\vec{\ell }$ is an elementary segment of this loop. This is also directed along the tangent.

Hence $\vec{B}\cdot d\vec{\ell }$ = Bd$\ell$

Here $\oint \vec{B}\cdot d\vec{\ell }$= line integral of $\vec{B}$over closed path.

i is the net current linked by the Amperes loop. From symmetry, the magnitude of the magnetic field at any point on the Amperes loop is always the same and so

$\oint \vec{B}\cdot d\vec{\ell }=B\oint d\ell$

$\Rightarrow B\oint d\ell ={\mu }_{o}i$

$\Rightarrow$ B.2$\pi$r =$\mu$_oi $\Rightarrow$ B = $\frac{{\mu }_{o}i}{2\pi r}$

Illustration 1: A current i flows along a long thin walled tube of radius r with a longitudinal slit of width h. Find the induction of magnetic field inside the tube (h <<r)

Solution: This tube can be considered as a complete tube and a thin strip of width h, carrying a current of equal density, in the opposite direction. Field due to the complete tube inside the tube is zero. So the magnetic field is due to the strip only

MAGNETIC FIELD B AT A POINT ON THE AXIS OF A LONG SOLENOID

A solenoid consists of a long wire wound closely in the form of a helix (spring). The gap between two consecutive turns is negligibly small and plane of each loop is perpendicular to the axis of the solenoid.

ABCD is a closed loop

Since $\oint \vec{B}\cdot d\vec{\ell }={\mu }_{o}i$ $\Rightarrow \underset{A}{\overset{B}{\int }}\vec{B}\cdot d\vec{\ell }+\underset{B}{\overset{C}{\int }}\vec{B}\cdot d\ell +\underset{C}{\overset{D}{\int }}\vec{B}\cdot d\ell +\underset{D}{\overset{A}{\int }}\vec{B}\cdot d\vec{\ell }={\mu }_{o}i$

Solenoid is the combination of circular loops. Due to ideal solenoid B is only along the axis and out side the solenoid B is zero. AB, and CD are perpendicular to $\overrightarrow{B}$ hence except $\underset{B}{\overset{C}{\int }}\vec{B}\cdot d\vec{\ell }$all integral is zero.

Hence $\underset{B}{\overset{C}{\int }}\vec{B}\cdot d\vec{\ell }={\mu }_{o}i\Rightarrow B\underset{B}{\overset{C}{\int }}d\ell ={\mu }_{o}i$

If number of turns per unit length of solenoid is n and length of BC is $\ell$ then i = n$\ell$i

And hence

B$\ell$ = $\mu$_on$\ell$i $\Rightarrow$ B = $\mu$_oni

This result is applicable for point in the central part of the solenoid.

Illustration 2: A solenoid switch is activated when the magnetic induction at the axis is 5 × 10 ^{– 4} Wb/m². The solenoid has 50 turns per cm. and an inductance of 180 mH and is operated by a 12V battery. Find the time lag when it is employed in a circuit of a resistance 90 ohm.

Solution: When the magnetic induction in the solenoid has value B, then current i passing through the coil is given by B = $\mu$₀n i

(n = number of turns per unit length in solenoid)

The current i' at any time t after the solenoid current is switched on is given by

$i^{\prime }=\frac{V}{R}\left[1-e^{-Rt/L}\right]$ (L = Solenoid inductance)

$B={\mu }_{0}n{i}^{\prime }=\frac{{\mu }_{0}nV}{R}\left[1-e^{-Rt/L}\right]$

or $\frac{BR}{{\mu }_{0}nV}=1-e^{-Rt/L}$

or $e^{-Rt/L}=1-\frac{BR}{{\mu }_{0}nV}$

–$\frac{Rt}{L}={\log }_e\left[1-\frac{BR}{{\mu }_{0}nV}\right]$

$t=-\frac{L}{R}{\log }_e\left[1-\frac{BR}{{\mu }_{0}nV}\right]$

Substituting the given values, we get

$t=-\frac{180\times {10}^{-3}}{90}{\log }_e\left[1-\frac{(5\times {10}^{-4})(90)}{(4\pi \times {10}^{-7})(5\times {10}^3)(12)}\right]$

= – 2 x 10 ^{– 3} x 2.303 log (1 – 0.597)

= 4.606 x 10 ^{– 3} log (1/ 0.403) = 1.818 x 10 ^{– 3} second.