Question

A long straight wire of radius a carries a steady current I. The current is uniformly distributed over its cross-section. The ratio of the magnetic fields a/2 and 2a is?
A) 1/4​
B) 4​
C) 1
D) 1/2

Answer 1

Here in this question we will use the concept of Ampere’s law which states that the magnetic field created by an electric current is proportional to the size of that electric current with a constant of proportionality equal to the permeability of free space.

Complete Step by step solution: Consider two ampere loops of radius a / 2 and 2a
Applying the ampere law of current for these loops, we obtain
\eqalign{ & \oint\limits_{} {{\text{B}}{\text{.dL = }}{\mu _o}} {\text{ }}{{\text{I}}_{{\text{enclosed}}}} \cr & {\text{For the smaller loop,}} \cr & \Rightarrow {\text{B }} \times {\text{ 2}}\pi \dfrac{a}{2}{\text{ = }}{\mu _o} \times {\text{ }}\dfrac{1}{{\pi {a^2}}}{\text{ }} \times {\text{ }}\dfrac{{{a^2}}}{{{2^2}}} \cr & \Rightarrow {\mu _o}{\text{I }} \times {\text{ }}\dfrac{1}{4}{\text{ = }}\dfrac{{{\mu _o}{\text{I}}}}{4} \cr & \Rightarrow {\text{B = }}\dfrac{{{\mu _o}{\text{I}}}}{{4\pi {\text{a}}}} \cr & \Rightarrow {\text{B' }} \times {\text{ 2}}\pi {\text{ }} \times {\text{ 2a = }}{\mu _o}{\text{I}} \cr & \Rightarrow \dfrac{{\text{B}}}{{{\text{B'}}}} = {\text{ }}\dfrac{{{\mu _o}{\text{I}}}}{{4\pi {\text{a}}}}{\text{ }} \times {\text{ }}\dfrac{{4\pi {\text{a}}}}{{{\mu _o}{\text{I}}}} \cr & \therefore {\text{ }}\dfrac{{\text{B}}}{{{\text{B'}}}} = 1 \cr}

Hence option C is correct

Additional Information: The equation for the magnetic field strength (magnitude) produced by a long straight current conducting wire is:
${\text{B = }}\dfrac{{{\mu _o}{\text{I}}}}{{2\pi {\text{r}}}}$
The magnetic field of a long straight cable has more implications than one might suspect. Each current segment produces a magnetic field like that of a long straight cable, and the total field of any current form is the vector sum of the fields due to each segment. The formal statement of the direction and magnitude of the field due to each segment is called the Biot-Savart law. Integral calculus is needed to add the field of a current arbitrarily.
Notes: The electric current produces a magnetic field. This magnetic field can be visualized as a pattern of circular field lines surrounding a wire. One way to explore the direction of a magnetic field is with a compass,by a long straight current conductor. Hall probes can determine the magnitude of the field.
Another version of the right hand rule emerges from this exploration and is valid for any current segment: When we point our thumb in the direction of the current and your fingers curve in the direction of the magnetic field loops created by it.