
The relation between magnetic field and current is given by Biot-Savart law. Illustrate Biot-Savart law with necessary figures.
Answer
567k+ views
Hint
The Biot-Savart law states that a small current-carrying conductor of length $ dl $ and carrying a current $ I $ is an elementary source of the magnetic field. Therefore this law can be used to calculate the magnetic field of certain distributions.
Complete step by step answer
The Biot-Savart law gives us the magnetic field that is associated with a current-carrying conductor. According to this law, the magnetic field at any point due to a current element $ Idl $ is given by,
$\Rightarrow d\vec B = \dfrac{{{\mu _o}}}{{4\pi }}\dfrac{{Id\vec l \times \vec R}}{{{R^3}}} $
$\vec R $ is the distance of the current-carrying conductor from the point of observation and $ {\mu _o} $ is the permeability of free space which has a value of $ {\mu _o} = 4\pi \times {10^{ - 7}}N/{A^2} $ .
To derive this mathematical expression we consider a wire carrying current $ I $ in a specific direction as in the figure,
Let us consider a small element of wire $ dl $ . The direction of this element is along the direction of the current in the wire. Now by using the Biot-Savart law we can calculate the magnetic field at the point P due to this current element.
The magnetic field at point P due to the element $ dl $ is found to be proportional to the current in the wire, and the length of $ dl $ and is inversely proportional to the square of the distance $ R $ .
So we can write,
$\Rightarrow d\vec B \propto \dfrac{{Id\vec l \times \vec R}}{{{R^3}}} $
and removing the proportionality we get,
$\Rightarrow d\vec B = \dfrac{{{\mu _o}}}{{4\pi }}\dfrac{{Id\vec l \times \vec R}}{{{R^3}}} $
The magnitude of this field is given by,
$\Rightarrow \left| {d\vec B} \right| = \dfrac{{{\mu _o}}}{{4\pi }}\dfrac{{Idl\sin \theta }}{{{R^2}}} $.
Note
The Biot-Savart law is similar to Coulomb's law in a way as both of them are inversely proportional to the square of the distance between the source and the point. Using the Biot-Savart law we can calculate the magnetic fields of various current-carrying elements.
The Biot-Savart law states that a small current-carrying conductor of length $ dl $ and carrying a current $ I $ is an elementary source of the magnetic field. Therefore this law can be used to calculate the magnetic field of certain distributions.
Complete step by step answer
The Biot-Savart law gives us the magnetic field that is associated with a current-carrying conductor. According to this law, the magnetic field at any point due to a current element $ Idl $ is given by,
$\Rightarrow d\vec B = \dfrac{{{\mu _o}}}{{4\pi }}\dfrac{{Id\vec l \times \vec R}}{{{R^3}}} $
$\vec R $ is the distance of the current-carrying conductor from the point of observation and $ {\mu _o} $ is the permeability of free space which has a value of $ {\mu _o} = 4\pi \times {10^{ - 7}}N/{A^2} $ .
To derive this mathematical expression we consider a wire carrying current $ I $ in a specific direction as in the figure,
Let us consider a small element of wire $ dl $ . The direction of this element is along the direction of the current in the wire. Now by using the Biot-Savart law we can calculate the magnetic field at the point P due to this current element.
The magnetic field at point P due to the element $ dl $ is found to be proportional to the current in the wire, and the length of $ dl $ and is inversely proportional to the square of the distance $ R $ .
So we can write,
$\Rightarrow d\vec B \propto \dfrac{{Id\vec l \times \vec R}}{{{R^3}}} $
and removing the proportionality we get,
$\Rightarrow d\vec B = \dfrac{{{\mu _o}}}{{4\pi }}\dfrac{{Id\vec l \times \vec R}}{{{R^3}}} $
The magnitude of this field is given by,
$\Rightarrow \left| {d\vec B} \right| = \dfrac{{{\mu _o}}}{{4\pi }}\dfrac{{Idl\sin \theta }}{{{R^2}}} $.
Note
The Biot-Savart law is similar to Coulomb's law in a way as both of them are inversely proportional to the square of the distance between the source and the point. Using the Biot-Savart law we can calculate the magnetic fields of various current-carrying elements.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Trending doubts
What are the major means of transport Explain each class 12 social science CBSE

Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

How much time does it take to bleed after eating p class 12 biology CBSE

Explain sex determination in humans with line diag class 12 biology CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

