Question

A long straight wire of radius a carries steady current I .The current is uniformly distributed across its cross section, The ratio of the magnetic field B and B’, at radial distance a/2 and 2a respectively, from the axis of wire is.
$\begin{align}
  & \text{A}\text{. }\dfrac{\text{1}}{\text{4}} \\
 & \text{B}\text{. }\dfrac{\text{1}}{\text{2}} \\
 & \text{C}\text{. 1} \\
 & \text{D}.\text{ }4 \\
\end{align}$

Answer 1

Hint: We should have the basic idea of Ampere circuital law. We will use this law to find magnetic fields at radial distance a/2 and 2a. Once we have found the magnetic field we will see that both the magnetic fields are equal hence we will obtain the ratio.

Formula used:
$\oint{\vec{B}.d\vec{l}={{\mu }_{{}^\circ }}{{I}_{en}}}$


Complete step by step answer:
A long straight wire of radius ‘a’ carries steady current I. also question whether “current is uniformly distributed.”
Now, magnetic field due to straight long wire at perpendicular distance from its axis is found by Ampere circuital law,
$\oint{\vec{B}.d\vec{l}={{\mu }_{{}^\circ }}{{I}_{en}}}$
Here, ${{i}_{en}}$ is currently enclosed by an amperian path.
Case 1: magnetic field at a/2:
Here,
$\begin{align}
  & {{i}_{en}}=I\dfrac{\pi {{\left( \dfrac{a}{2} \right)}^{2}}}{\pi {{a}^{2}}}=\dfrac{I}{4} \\
 & \\
\end{align}$
So, $\vec{B}.d\vec{l}={{\mu }_{{}^\circ }}{{i}_{en}}$
$\begin{align}
  & B.2\pi \left( \dfrac{a}{2} \right)={{\mu }_{{}^\circ }}\dfrac{I}{4} \\
 & =B=\dfrac{{{\mu }_{{}^\circ }}I}{4\pi a} \\
\end{align}$
Case 2 : magnetic field at 2a :
Here, ${{i}_{en}}=I$ (because total current flows through wire is I.)
So $\vec{B}'.d\vec{l}'={{\mu }_{{}^\circ }}{{i}_{en}}$
$\begin{align}
  & B'.2\pi \left( 2a \right)={{\mu }_{{}^\circ }}I \\
 & =B'=\dfrac{{{\mu }_{{}^\circ }}I}{4\pi a} \\
\end{align}$
From both cases we obtain ratio
$\dfrac{B}{B'}=\dfrac{\dfrac{{{\mu }_{{}^\circ }}I}{4\pi a}}{\dfrac{{{\mu }_{{}^\circ }}I}{4\pi a}}=1$
The ratio of the magnetic field B and B’, at radial distance a/2 and 2a respectively, from the axis of wire is 1.
Therefore the correct option is C.

Note:
Ampere circuital law basically tells us about the magnetic field produced by electric current.
Biot-Savart law is analogous to Coulomb's law, where from the position and value of the currents, the magnetic flux value is often calculated. The Ampere’s law has two forms:
 The integral form which relates the line integral of the magnetic field over a closed loop to the current passing through the loop, here given the current passing through the loop one can calculate the line integral or vice versa.
 The differential form which gives the curl of the magnetic field (a kind of vector derivative) at a point in terms of the current density of that point.
The integral sort of the Ampere’s law are often want to find the worth of magnetic fields in space but just for certain symmetric configurations of currents, whereas Biot-Savart law can be wont to find the magnetic flux for any given current configuration.