Question

$AB$ and $CD$ are long straight conductors, distance $d$ apart, carrying a current $I$. The magnetic field at the
the midpoint of $BC$ is

• A. $\frac{-{\mu }_{0}I}{2\mathrm{\pi d}}\hat{k}$
• B. $\begin{array}{l}\frac{-{\mu }_{0}I}{\pi d}\hat{k}\end{array}$
• C. $\begin{array}{l}\frac{-{\mu }_{0}I}{4\pi d}\hat{k}\end{array}$
• D. $\begin{array}{l}\frac{-{\mu }_{0}I}{8\pi d}\hat{k}\end{array}$

Answer 1

The correct option is B

$\begin{array}{l}\frac{-{\mu }_{0}I}{\pi d}\hat{k}\end{array}$

Step 1: Given Data

Distance between the two conductors $a=d$

Therefore, the distance at the midpoint of $BC=\frac{d}{2}$

Current through both the conductors are equal $=I$

Let ${\mu }_0$ be the permeability of free space.

Step 2: Calculate the required magnetic field

From the figure, we can see that $AB$ and $CD$ will generate the same magnetic field at all points on $BC$ in the same direction.

Therefore, the magnitude of the magnetic field at the center of $BC$ can be given as,

$B=\frac{{\mu }_0}{4\mathrm{\pi }}\frac{2I}{a}$

$\Rightarrow B=\frac{{\mu }_0}{4\mathrm{\pi }}\frac{2I\times 2}{d}$

$\Rightarrow B=\frac{{\mu }_{0}I}{\mathrm{\pi d}}$

Step 3: Determine the Direction

1. According to Fleming's right-hand rule, the thumb represents the direction of motion, the index finger represents the direction of the magnetic field, and the middle finger represents the direction of the flow of current, such that all three of them are mutually perpendicular to each other.
2. Upon aligning the direction of the current from the given figure, we can see that if the current $I$ is in positive $\hat{i}$ direction, then the magnetic field $B$ comes to be in negative $\hat{k}$ direction.

Therefore, the magnetic field vector can be written as,

$\overrightarrow{B}=-\frac{{\mu }_{0}I}{\mathrm{\pi d}}\hat{k}$

Hence, the correct answer is option (B).