Question

The magnetic field (dB) due to smaller element (dl) at a distance $\overset{\to }{\mathop{r}}\,$ from element carrying current i, is

& A.\,dB=\dfrac{{{\mu }_{0}}i}{4\pi }\left( \dfrac{\overset{\to }{\mathop{dl}}\,\times \overset{\to }{\mathop{r}}\,}{r} \right) \\\ & B.\,dB=\dfrac{{{\mu }_{0}}i}{4\pi }{{i}^{2}}\left( \dfrac{\overset{\to }{\mathop{dl}}\,\times \overset{\to }{\mathop{r}}\,}{{{r}^{2}}} \right) \\\ & C.\,dB=\dfrac{{{\mu }_{0}}i}{4\pi }{{i}^{3}}\left( \dfrac{\overset{\to }{\mathop{dl}}\,\times \overset{\to }{\mathop{r}}\,}{2{{r}^{2}}} \right) \\\ & D.\,dB=\dfrac{{{\mu }_{0}}}{4\pi }i\left( \dfrac{\overset{\to }{\mathop{dl}}\,\times \overset{\to }{\mathop{r}}\,}{{{r}^{3}}} \right) \\\ \end{aligned}$$

Answer 1

The question is based on the concept of Biot-Savart’s law. This law gives the equation of the magnetic field produced due to the current-carrying element. This law is applied for the symmetrical current distribution and for the conductors of small size carrying current.

Complete step by step answer:
From the given information, we have the data as follows.
The magnetic field (dB) due to smaller element (dl) at a distance $\overset{\to }{\mathop{r}}\,$from element carrying current (i).
Biot-Savart law is used to compute the magnetic responses, even at a very small range. This law is similar to that of Coulomb’s law of electrostatics.
According to Biot-Savart law,
The small area of the magnetic field is directly proportional to the flow of current. $dB\propto i$
The small area of the change in a magnetic field is directly proportional to the vector distance. $dB\propto r$
The small area of the change in a magnetic field is directly proportional to the small current element. $dB\propto dl$
The small area of the change in a magnetic field is inversely proportional to the cube of the distance. $dB\propto \dfrac{1}{{{r}^{3}}}$

Combining all the above proportionality equations, we get, $dB\propto i\left( \dfrac{\overset{\to }{\mathop{dl}}\,\times \overset{\to }{\mathop{r}}\,}{{{r}^{3}}} \right)$
To remove this proportionality constant, we use, $\dfrac{{{\mu }_{0}}}{4\pi }$.
$dB=\dfrac{{{\mu }_{0}}}{4\pi }i\left( \dfrac{\overset{\to }{\mathop{dl}}\,\times \overset{\to }{\mathop{r}}\,}{{{r}^{3}}} \right)$
The value of the constant, magnetic permeability of free space, is given to be, ${{\mu }_{0}}=4\pi \times {{10}^{-7}}{H}/{m}\;$
$\therefore$ The magnetic field (dB) due to smaller element (dl) at a distance $\overset{\to }{\mathop{r}}\,$from element carrying current i, is $dB=\dfrac{{{\mu }_{0}}}{4\pi }i\left( \dfrac{\overset{\to }{\mathop{dl}}\,\times \overset{\to }{\mathop{r}}\,}{{{r}^{3}}} \right)$
So, the correct answer is “Option D”.

Note: The direction of the magnetic field is perpendicular to the plane containing the small current element and the distance. The magnetic field is directed inward. The current element is a vector quantity. Biot-Savart’s law gives the equation of the magnetic field produced due to the current carrying element.