Question

A small current element ${{Id\vec l}}$, with ${\text{d\vec l}} = 2\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{k} {\text{ }}mm$ and ${\text{I}} = 2{\text{A}}$ is centred at the origin. Find magnetic field $d\vec B$ āt the following points:
(A) On the x-axis at $x = 3m$
(B) On the x-axis at $x = 6m$
(C) On the z-axis at $z = 3m$.

Answer 1

Hint: To solve this question, we need to use the vector form of the Biot Savart’s law. For this we need to substitute the position vectors of the points given in all the three parts of this question to find out the magnitude and direction of the magnetic field in the respective cases.
Formula used: The formula used to solve this question is given by
\[d\vec B = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{Id\vec l \times \hat r}}{{{r^2}}}\], here \[d\vec B\] is the magnetic field produced by a current element \[Id\vec l\] at a point whose position vector is \[\vec r\]-with respect to the centre of the current element.

Complete step-by-step solution:
We know from the Biot Savart’s law that the magnetic field produced by a current element at a point is given by
\[d\vec B = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{Id\vec l \times \hat r}}{{{r^2}}}\] ………………………..(1)
Now, according to the question, we have ${\text{d\vec l}} = 2\hat k{\text{ }}mm = 2 \times {10^{ - 3}}\hat k{\text{ }}mm$ and ${\text{I}} = 2{\text{A}}$. Also we know that \[\dfrac{{{\mu _0}}}{{4\pi }} = {10^{ - 7}}Tm/A\]. Substituting these in (1) we get
\[d\vec B = {10^{ - 7}}\dfrac{{2 \times 2 \times {{10}^{ - 3}}\left( {\hat k \times \hat r} \right)}}{{{r^2}}}\]
$ \Rightarrow d\vec B = 4 \times {10^{ - 10}}\dfrac{{\left( {\hat k \times \hat r} \right)}}{{{r^2}}}$............................(2)
We have to determine the magnetic field on the x-axis at $x = 3m$. The position vector of this point can be given by $\vec r = 3\hat i{\text{ }}m$. This means that $\hat r = \hat i$ and $r = 3m$. Substituting these in (2) we get
$d\vec B = 4 \times {10^{ - 10}}\dfrac{{\left( {\hat k \times \hat i} \right)}}{{{3^2}}}$
On solving we get
\[d\vec B = 4.44 \times {10^{ - 11}}\hat j\]
We have to determine the magnetic field on the x-axis at $x = 6m$. The position vector of this point can be given by $\vec r = 6\hat i{\text{ }}m$. This means that $\hat r = \hat i$ and $r = 6m$. Substituting these in (2) we get
$d\vec B = 4 \times {10^{ - 10}}\dfrac{{\left( {\hat k \times \hat i} \right)}}{{{6^2}}}$
On solving we get
\[d\vec B = 1.11 \times {10^{ - 11}}\hat j\]
We have to determine the magnetic field on the z-axis at $z = 3m$. The position vector of this point can be given by $\vec r = 3\hat k{\text{ }}m$. This means that $\hat r = \hat k$ and $r = 3m$. Substituting these in (2) we get
$d\vec B = 4 \times {10^{ - 10}}\dfrac{{\left( {\hat k \times \hat k} \right)}}{{{3^2}}}$
On solving we get
\[d\vec B = 0\]

Note: Prefer to use the vector form of the Biot Savart’s law to solve these types of questions. This is because otherwise we have to calculate the angle between the position vector and the current element, which will make the solution complex.