Question

The Biot Savart’s law in vector form is
A. \[\overline{\delta B}=\dfrac{{{\mu }_{o}}}{4\pi }\dfrac{dl(\overrightarrow{l}\times \overrightarrow{r})}{{{r}^{3}}}\]
B. \[\overline{\delta B}=\dfrac{{{\mu }_{o}}}{4\pi }\dfrac{I(\overrightarrow{dl}\times \overrightarrow{r})}{{{r}^{3}}}\]
C. \[\overline{\delta B}=\dfrac{{{\mu }_{o}}}{4\pi }\dfrac{I(\overrightarrow{r}\times \overrightarrow{dl})}{{{r}^{3}}}\]
D. \[\overline{\delta B}=\dfrac{{{\mu }_{o}}}{4\pi }\dfrac{I(\overrightarrow{dl}\times \overrightarrow{r})}{{{r}^{2}}}\]

Answer 1

Hint: The magnitude of the magnetic field at any point is proportional to the current I, the element length dl and inversely proportional to the square of the distance r of the current element from the point.

Complete step by step solution:
According to Biot-Savart’s law, the magnetic field at any point due to a current element \[I\overrightarrow{dl}\] is given by:
\[\overline{\delta B}=\dfrac{{{\mu }_{o}}}{4\pi }\dfrac{Idl\sin \theta }{{{r}^{2}}}\overset{\wedge }{\mathop{r}}\,\]
Where \[\theta \] is the angle between the length element \[\overrightarrow{dl}\] and \[\overrightarrow{r}\] , which is the displacement vector from the current element to the point; and \[\overset{\wedge }{\mathop{r}}\,\] is the displacement unit vector. The constant \[{{\mu }_{o}}\]is called the permeability of free space.
The unit vector \[\overset{\wedge }{\mathop{r}}\,\] is the ratio of displacement vector \[\overrightarrow{r}\] and the magnitude of displacement r.
\[\overset{\wedge }{\mathop{r}}\,=\dfrac{\overrightarrow{r}}{r}\]
Also, cross product of the vectors \[\overrightarrow{dl}\times \overrightarrow{r}=dlr\sin \theta \]

So, the vector form of Biot Savart’s law can be written as
\[\overline{\delta B}=\dfrac{{{\mu }_{o}}}{4\pi }\dfrac{I(\overrightarrow{dl}\times \overrightarrow{r})}{{{r}^{3}}}\]

So, option B is the correct answer.

Additional information:
The direction of the magnetic field will be the direction of the cross product \[\overrightarrow{dl}\times \overrightarrow{r}\]. It is represented by the right hand rule. The magnetic field inside the conductor is zero, while on the surface, the magnetic field is perpendicular to both the current density and the surface normal.
From Biot Savart’s law, it can be shown that the magnetic field due to the current carrying conductor depends on the current flowing through the conductor and the distance from the conductor.
The magnetic field at any point due to a current carrying wire is
\[\overset{\to }{\mathop{B}}\,=\dfrac{{{\mu }_{o}}}{4\pi }\int{\dfrac{I\overrightarrow{dl}\times \overset{\to }{\mathop{r}}\,}{{{r}^{3}}}}\]
Where the integration is over the whole current path.

Note: The Biot-Savart law is applicable only for steady currents.
The cross product of any two vectors is not commutative.
So, \[\overrightarrow{r}\times \overrightarrow{dl}\ne \overrightarrow{dl}\times \overrightarrow{r}\].
The angle \[\theta \] is the angle between the length element \[\overrightarrow{dl}\] and \[\overrightarrow{r}\].