Question

A straight wire of a diameter $0.5\,mm$ , carrying current of $1\,A$ is replaced by another wire of $1\,mm$ diameter carrying the same current. The strength of magnetic field far away is
A. Unchanged
B. Quarter of its earlier value
C. Half of the earlier value
D. Twice the earlier value

Answer 1

A magnetic field is produced by moving electric charges and intrinsic magnetic moments of particles associated with a fundamental quantum property of spin. The magnetic field at a distance $r$ from a long current-carrying conductor carrying current $I\,A$ is given by the equation $B\,=\,\dfrac{{{\mu }_{o}}I}{2\pi r}$ .

Complete step by step solution:
The magnetic field of a straight long current-carrying conductor, carrying current $I\,A$ at a distance $r$ is given by
$B\,=\,\dfrac{{{\mu }_{o}}I}{2\pi r}$
Here,
Magnetic flux density, $B$ is measured in $T$ (Tesla).
Currently, $I$ is measured in $A$ (Ampere).
The distance from the current carrying conductor is given by, $r$ in $m$ (meters).
The permeability of free space is given by, ${{\mu }_{o}}\,=\,4\pi \,\times \,{{10}^{-7\,}}\,$ in ${N}/{{{A}^{2}}}\;$ (newton per ampere square).
Here, $\dfrac{{{\mu }_{o}}}{2\pi }$ is a constant quantity
Therefore, $B\,\alpha \dfrac{I}{r}$
Hence, we can see that
The strength of the magnetic field depends on the current, $I$ and the distance, $r$ .
The strength of the magnetic field is directly proportional to the current, $I$ of the straight current-carrying conductor and is inversely proportional to the distance, $r$ from the straight current-carrying conductor.
Hence, we can say that as the current through the current-carrying conductor increases, the magnetic field increases proportionally. When we move further away from the conductor, the magnetic field decreases with the distance.
Therefore, the magnetic field does not depend on the radius or diameter of the current-carrying conductor. Changing the radius or diameter of the current-carrying conducting wire will not have any effect on the strength of the magnetic field.
Now, according to the question:
Wire (i) has a diameter of $0.5\,mm$ and current, $I\,=\,1\,A$ .
Wire (ii) has a diameter of $1\,mm$ and current, $I\,=\,1\,A$ .
Since the strength of the magnetic field does not depend on the diameter of the wire and current for both wires, (i) and (ii) are the same.
Hence, there is no change in the strength of the magnetic field. So, the strength of the magnetic field remains unchanged.
Therefore, option (A) unchanged is the correct option.

Note:
The strength of the magnetic field due to a straight current-carrying conductor can be parted into two different cases, case(i) when the length of wire is finite in this case the strength of the magnetic field is given by $B\,=\,\dfrac{{{\mu }_{o}}I}{4\pi r}\left( \sin {{\phi }_{1}}\,+\,\sin {{\phi }_{2}} \right)$ and in case(ii) when the length of wire is infinite in this case the strength of the magnetic field is given by $B\,=\,\dfrac{{{\mu }_{o}}I}{2\pi r}$ .