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Question: A long straight wire of circular cross section of radius ‘a’ carries a steady current \(I\). The cur...

A long straight wire of circular cross section of radius ‘a’ carries a steady current II. The current is uniformly distributed across the cross-section .Apply Ampere’ circular law to calculate the magnetic field at a point ‘r’ in the region for (i) rar \prec a and (ii) rar \succ a.Plot a graph showing the nature of variation.

Explanation

Solution

Before going through the question let us first discuss Magnetism. Magnetism is a type of physical phenomenon that is mediated by magnetic fields. Electric currents and the magnetic moments of elementary particles combine to form a magnetic field, which interacts with other currents and magnetic moments. Electromagnetism is a multifaceted phenomenon that includes magnetism.

Complete answer:
Ampere’s Circuital law states that: According to Ampere's circuital theorem, the line integral of the magnetic field surrounding a closed-loop equals the algebraic number of currents passing through the loop.
HdL=Ience\oint {H \cdot dL = {I_{ence}}}
If a conductor is carrying current I, the current flow creates a magnetic field around the cable.
The left side of the equation states that if a magnetic field is applied at each point along an imaginary path that encircles the wire, the current encircled by this path, as indicated by Ienc, is numerically equal to the current encircled by this route.
Now, coming to the question,

(i) considering the case rar \prec a. The Ampere loop is a circle that is labelled as 1. For this loop , let’s take the radius of the as ‘r’, L=2πrL = 2\pi r.
The new enclosed current Ie{I_e} is no longer equal toII, but it is now less than this amount. The latest current enclosed is, since the current distribution is uniform.
Ie=I(πr2πa2)=Ir2a2{I_e} = I\left( {\dfrac{{\pi {r^2}}}{{\pi {a^2}}}} \right) = \dfrac{{I{r^2}}}{{{a^2}}}
Using Ampere's law :
B(2πr)=μIr2a2 B=(μI2πa2)r Br(ra)  B\left( {2\pi r} \right) = {\mu _ \circ } \cdot \dfrac{{I{r^2}}}{{{a^2}}} \\\ B = \left( {\dfrac{{{\mu _ \circ }I}}{{2\pi {a^2}}}} \right)r \\\ B \propto r\,\,\,\,\,\,\,(r \prec a) \\\
(ii) Now, considering the case rar \succ a
The Amperian loop, denoted by the number 2, is a circle that is concentric to the cross-section. For this loop,
L=2πrL = 2\pi r
Ie{I_e} = Enclosed Current by the loop II
As a consequence, we have the well-known term for a long straight wire.
B(2πr)=μI B=μI2πr B1r(ra)  B\left( {2\pi r} \right) = {\mu _ \circ }I \\\ B = \dfrac{{{\mu _ \circ }I}}{{2\pi r}} \\\ B \propto \dfrac{1}{r}\,\,\left( {r \succ a} \right) \\\
The magnitude of B is plotted against the distance r from the wire's centre in the figure. The right-hand direction determines the area, which is tangential to the respective circular loop (1 or 2).

The magnitude of B is plotted against the distance r from the wire's centre in the figure. The right-hand direction defines the field, which is tangential to the respective circular loop (1 or 2).

Note:
Ampere's Law is a mathematical relationship between magnetic fields and electric currents that helps us to bridge the difference between electricity and magnetism. It allows us to measure the magnetic field generated by an electric current flowing through any form of wire.