#### Current loop as a Magnetic Dipole

Ampere found that the distribution of magnetic lines of force around a finite current carrying solenoid is similar to that produced by a bar magnet. This is evident from the fact that a compass needle when moved around these two bodies show similar deflections. After noting the close resemblance between these two, Ampere demonstrated that a simple current loop behaves like a bar magnet and put forward that all the magnetic phenomena is due to circulating electric current. This is Ampere’s hypothesis. We consider a circular coil carrying current I. When seen from above current flows in anti clockwise direction.- The magnetic field lines due to each elementary portion of the circular coil are circular near the element and almost straight near center of circular coil.
- The magnetic lines of force seem to enter at lower face of coil and leave at upper face.
- The lower face through which lines of force enter behaves as south pole and upper face through which field lines leave behaves as north pole.
- A planar loop of any shape behaves as a magnetic dipole.
- The dipole moment of current loop $(\mathrm{M})=$ ampere turns (nI) $\times$ area of coil (A) or $\mathrm{M}=\mathrm{nIA}$.
- The unit of dipole moment is ampere meter $^{2}\left( A – m ^{2}\right)$
- Magnetic dipole moment is a vector with direction from S pole to N pole or along direction of normal to planar area.

### Atoms as a Magnetic Dipole

In an atom electrons revolve around the nucleus. These moving electrons behave as small current loops. So atom possesses magnetic dipole moment and hence behaves as a magnetic dipole. The angular momentum of electron due to orbital motion $L = m _{ e } vr$ The equivalent current due to orbital motion $I =-\frac{ e }{ T }=-\frac{ ev }{2 \pi r }$ –ve sign shows direction of current is opposite to direction of motion of electron. Magnetic dipole moment $M=I A=-\frac{e v}{2 \pi r} \cdot \pi r^{2}=-\frac{e v r}{2}$ Using $L=m_{e}$ vr we have $\quad M=-\frac{e}{2 m_{e}} L$ In vector form $\overrightarrow{ M }=-\frac{ e }{2 m _{ e }} \overrightarrow{ L }$ The direction of magnetic dipole moment vector is opposite to angular momentum vector. According to Bohr’s theory $L=\frac{n h}{2 \pi} \quad n=0, \quad 1, \quad 2 \ldots \ldots$ So $M=\left(\frac{e}{2 m_{e}}\right) \frac{n h}{2 \pi}=n\left(\frac{e h}{4 \pi m_{e}}\right)=n \mu_{B}$ Where $\mu_{ B }=\frac{ eh }{4 \pi m _{ e }}=\frac{\left(1.6 \times 10^{-19} C \right)\left(6.62 \times 10^{-34} Js \right)}{4 \times 3.14 \times\left(9.1 \times 10^{-31} kg \right)}=9.27 \times 10^{-24} Am ^{2}$ is called**Bohr Magneton**. This is natural unit of magnetic moment.

**Also Read:**Biot Savart’s Law

**Click here for the Video tutorials of Magnetic Effect of Current Class 12**

For free video lectures and complete study material, DownloadAbout eSaralAt eSaral we are offering a complete platform for IIT-JEE & NEET preparation. The main mission behind eSaral is to provide education to each and every student in India by eliminating the Geographic and Economic factors, as a nation’s progress and development depends on the availability of quality education to each and every one. With the blend of education & technology, eSaral team made the learning personalized & adaptive for everyone.

**eSaral APP.**