Question

A long wire with a small current element of length $1cm$ is placed at the origin and carries a current of $10A$ along the X-axis. Find out the magnitude and direction of the magnetic field due to the element on the Y-axis at a distance of $0.5cm$ from it.

Answer 1

Hint: This problem can be solved by drawing a proper figure for the problem and applying the formula for the magnetic field produced at a point due to a straight current carrying conductor. The magnitude can be found by this formula and the direction of the magnetic field is given by the rule for cross product of vectors.

Formula used:
The magnetic field $B$ produced at a point at a perpendicular distance $R$ by a long straight current carrying wire of length $L$ and carrying current $I$ is given by,
$\overrightarrow{B}=\dfrac{{{\mu }_{0}}I}{4\pi R}\left( \sin {{\phi }_{1}}+\sin {{\phi }_{2}} \right)\left( \widehat{L}\times \widehat{R} \right)$
where ${{\phi }_{1}}$ and ${{\phi }_{2}}$ are the angles made by the lines joining each respective end of the wire with the point with the line perpendicular to the wire and joining the point with the midpoint of the wire.
The direction of the magnetic field is such that it is perpendicular to the plane of the wire and the line joining the midpoint of the wire to the point. It is given by the cross product direction of $\left( \widehat{L}\times \widehat{R} \right)$

Complete step by step answer:
We will solve this problem by applying the formula for the magnetic field produced at a point due to a straight current carrying conductor.
First, let us draw a proper coordinate diagram.

Now, let us analyze the question.
Let unit vectors in the positive X, Y and Z axes be $\widehat{i}$,$\widehat{j}$ and $\widehat{k}$ respectively.
Now, we will take the length to be in the direction of the flow of the current.
Therefore length vector of the wire $\overrightarrow{L}=L\widehat{i}$ --(1)
$L=1cm=0.01m$
$R=0.5cm=0.005m$
The distance vector $\overrightarrow{R}=R\widehat{k}=0.5\widehat{k}$ --(2)
Current carried by the wire $I=10A$.
Now, as we can see from the figure,
${{\phi }_{1}}={{\phi }_{2}}={{\tan }^{-1}}\dfrac{\dfrac{L}{2}}{R}={{\tan }^{-1}}\left( \dfrac{L}{2R} \right)={{\tan }^{-1}}\left( \dfrac{0.01}{2\times 0.005} \right)={{\tan }^{-1}}\left( 1 \right)={{45}^{0}}$
The magnetic field $B$ produced at a point at a perpendicular distance $R$ by a long straight current carrying wire of length $L$ and carrying current $I$ is given by,
$\overrightarrow{B}=\dfrac{{{\mu }_{0}}I}{4\pi R}\left( \sin {{\phi }_{1}}+\sin {{\phi }_{2}} \right)\left( \widehat{L}\times \widehat{R} \right)$ --(3)
where ${{\phi }_{1}}$ and ${{\phi }_{2}}$ are the angles made by the lines joining each respective end of the wire with the point with the line perpendicular to the wire and joining the point with the midpoint of the wire.
The direction of the magnetic field is such that it is perpendicular to the plane of the wire and the line joining the midpoint of the wire to the point. It is given by the cross product direction of $\left( \widehat{L}\times \widehat{R} \right)$
Now, using (1), we get
$B=\dfrac{{{\mu }_{0}}10}{4\pi \times 0.005}\left( \sin {{45}^{0}}+\sin {{45}^{0}} \right)=\dfrac{{{10}^{-7}}\times 10}{5\times {{10}^{-3}}}\times \left( 2\times \dfrac{1}{\sqrt{2}} \right)=2\sqrt{2}\times {{10}^{-4}}\approx 2.83\times {{10}^{-4}}T$
Therefore, the magnitude of the magnetic field is $2.83\times {{10}^{-4}}T$.
The direction of the magnetic field is given by,
$\left( \widehat{L}\times \widehat{R} \right)$
Now, using (1) and (2), we get,
$\widehat{B}=\widehat{i}\times \widehat{j}=\widehat{k}$
Hence, the magnetic field will be in the positive Z axis.

Note: Students must be careful of marking the proper angles while drawing a diagram. Such types of questions are very specific in their directions and specific angles have to be implemented since vectors are being used. A very common mistake is that the length vector of the wire is taken as the opposite of the direction of the current.