Electrostatic analog class 12 | Relation between electrostatic and magnetism

Knowledge Increases By Sharing...
Electrostatic analog and the relation between electrostatic and magnetism

In this article, we are going to discuss the analogy of electrostatic and magnetism, so let’s get started…

Electrostatic and magnetism are complementary to each other. Wherever the current exists then there must exist Magnetism because magnetism is the effect of moving current. We have seen that wherever a current flows in the wire there exists a magnetic field and we have derived that what will the magnetic field in this case at some distance r from the current-carrying straight wire.

Suggested reading:

Electrostatic analog

In electrostatic, we have studied the electric field, electric dipole moment, electrostatic potential, torque on the electric dipole, the electric field at the axial and equatorial point of an electric dipole, and electrostatic potential energy, and many more things.[latexpage]

But when we study magnetism, then we feel that all these things are the same just some notations are changed. In Magnetism, definitions and formulae of various quantities look similar to what we have studied in electrostatic. For example: Electric field is given by $$E =\frac{F}{q_e}$$ where $q_e$ is electric charge. Similarly, the magnetic field can be given as $$B=\frac{F}{q_m}$$ where $q_m$ is the magnetic charge.

Suggested reading:

The magnetic dipole moment of a bar magnet is given as $$M=m(2l)$$ where $m$ is the strength of each pole and $2l$ is the length between the poles.

The magnetic dipole is analogous to an electric dipole consisting of two equal charges of opposite sign $(\pm q)$ seperated by some distance $(2a)$. The electric dipole moment can be given as: $$p=q(2a)$$

The equation for the magnetic field $(B)$ due to a magnetic dipole can be obtained from the equation of electric field $(E)$ due to an electric dipole by making the following changes. $$E \rightarrow B$$ $$p\rightarrow M$$ $$\frac{1}{4\pi\epsilon_0}\rightarrow\frac{\mu_0}{4\pi}$$

Thus, for any point on the axial line of a bar magnet at a distance $d(d>>>>l)$ from the centre of the magnet, magnetic field is given as: $$B_{axial}=\frac{\mu_0}{4\pi}\frac{2M}{d^3}$$ Similarly, equatorial field $(B_{equa})$ of a bar magnet at a distance $d(d>>>>l)$ is given as: $$B_{equa}=\frac{\mu_0M}{4\pi d^3}$$

Analogy between electrostatic and magnetism are given below in the table.

QuantityElectrostaticsMagnetism
Constant$\frac{1}{4\pi\epsilon_0}$ $\frac{\mu_0}{4\pi}$
Dipole moment$p$$M$
Equatorial field for short dipole $\frac{-p}{4\pi\epsilon_0. r^3}$$\frac{-\mu_0M}{4\pi r^3}$
Axial field for short dipole$\frac{1}{4\pi\epsilon_0}\frac{2p}{r^3}$$\frac{\mu_0}{4\pi}\frac{2M}{r^3}$
External field: Torque$ p\times E $$M\times B$
External field: energy$-p\cdot E$$-M\cdot B$

Stay tuned with Laws Of Nature for more useful and interesting content.

Was this article helpful?
YesNo
Knowledge Increases By Sharing...

Newsletter Updates

Enter your email address below and subscribe to our newsletter

0 0 votes
Rate this Article
Subscribe
Notify of
guest

0 Comments
Inline Feedbacks
View all comments
0
Would love your thoughts, please comment.x
()
x