As you know that the science is filled with fun facts. The deeper one dives into the concepts of science and its related fields, the greater amount of knowledge and information there is to learn in there. One such topic of study is the Gauss Law, which studies electric Charge along with a surface and the topic of Electric Flux. Let us study about the Gauss Law definition , Formula, Solved Examples in this Article,.

Gauss’s law states that the net flux of an Electric Field in a closed surface is directly proportional to the enclosed electric charge. It is one of the four equations of Maxwell’s laws of electromagnetism. It was initially formulated by Carl Friedrich Gauss in the year 1835 and relates the electric fields at the points on a closed surface and the net charge enclosed by that surface.

The surface integral of magnetic field $\overrightarrow{ B }$ over a closed surface S is always zero Mathematically $\oint_{S} \vec{B} \cdot \overrightarrow{d a}=0$

- Isolated magnetic poles do not exist is a direct consequence of gauss law in magnetism.
- The total magnetic flux linked with a closed surface is always zero.
- If a number of magnetic field lines are leaving a closed surface, an equal number of field lines must also be entering the surface.

**Ex. A bar magnet of length 0.1 m has a pole strength of 50 Am. Calculate the magnetic field at a distance of 0.2 m from its centre on its equatorial line.**

**Sol. **$B _{ equi }=\frac{\mu_{0}}{4 \pi} \frac{ M }{\left( r ^{2}+\ell^{2}\right)^{\frac{3}{2}}}=\frac{10^{-7} \times 50 \times 0.1}{\left(0.2^{2}+0.05^{2}\right)^{\frac{3}{2}}}=\frac{5 \times 10^{-7}}{(0.04+0.0025)^{\frac{3}{2}}}$ or $B _{\text {equi }}=5.7 \times 10^{-5}$ Tesla

**Ex. What is the magnitude of the equatorial and axial fields due to a bar magnet of length 5 cm at a distance of 50 cm from its mid-point. The magnetic moment of the bar magnet is 0.40 $Am ^{2}$**

**Sol. **Here r >> $\ell$. So equatorial field $B _{\text {equi }}=\frac{\mu_{0}}{4 \pi} \frac{ M }{ r ^{3}}=\frac{10^{-7} \times 0.4}{(0.5)^{3}}=3.2 \times 10^{-7} T$

Axial field $B _{\text {axial }}=\frac{\mu_{0}}{4 \pi} \frac{2 M }{ r ^{3}}=2 \times 3.2 \times 10^{-7}=6.4 \times 10^{-7} T$

**Ex. Find the magnetic field due to a dipole of magnetic moment 1.2 $Am ^{2}$ at a point 1 m away from it in a direction making an angle of 60° with the dipole axis?**

**Sol.** $B =\frac{\mu_{0}}{4 \pi} \frac{ M }{ r ^{3}} \sqrt{1+3 \cos ^{2} \theta}=\frac{10^{-7} \times 1.2 \sqrt{1+3 \cos ^{2} 60}}{1}=\frac{10^{-7} \times 1.2 \times \sqrt{7}}{2}=1.59 \times 10^{-7} T$

$\tan \theta^{\prime}=\frac{1}{2} \tan \theta=\frac{1}{2} \tan 60^{\circ}=\frac{\sqrt{3}}{2}=0.866$

So $\theta^{\prime}=\tan ^{-1} 0.866=40.89^{\circ}$

**Ex. A circular coil of 100 turns and having a radius of 0.05 m carries a current of 0.1 A. Calculate the work required to turn the coil in an external field of 1.5 T through $180^{\circ}$ about an axis perpendicular to the magnetic field. The plane of coil is initially at right angles to magnetic field?**

**Sol. **Work done W = MB

$\left(\cos \theta_{1}-\cos \theta_{2}\right)=N I A B\left(\cos \theta_{1}-\cos \theta_{2}\right)$

or $W = NI _{ B }^{2} B \left(\cos \theta_{1}-\cos \theta_{2}\right)=100 \times 0.1 \times 3.14 \times(0.05)^{2} \times 1.5\left(\cos 0^{\circ}-\cos \pi\right)=0.2355 J$

**Ex. A bar magnet of magnetic moment $1.5 HT ^{-1}$ lies aligned with the direction of a uniform magnetic field of 0.22 T. (a) What is the amount of work required to turn the magnet so as to align its magnetic moment. (i) Normal to the field direction? (ii) Opposite to the field direction? (b) What is the torque on the magnet in case (i) and (ii)?**

**Sol.** Here, M = $1.5 JT ^{-1}, B =0.22 T$

(a) P.E. with magnetic moment aligned to field = – MB

P.E. with magnetic moment normal to field = 0

P.E. with magnetic moment antiparallel to field = + MB

(i) Work done = increase in P.E. = 0 – (–MB) = MB = 1.5 × 0.22 = 0.33 J.

(ii) Work done = increase in P.E. = MB – (–MB) = 2MB = 2 × 1.5 × 0.22 = 0.66 J.

(b) We have $\tau$ = MB sin $\theta$

(i) $\theta=90^{\circ}, \sin \theta=1, \tau= MB \sin \theta=1.5 \times 0.22 \times 1=0.33 J$

This torque will tend to align M with B.

(ii) $\theta=180^{\circ}, \sin \theta=0, \tau= MB \sin \theta=1.5 \times 0.22 \times 0=0$

**Ex. A short bar magnet of magnetic moment 0.32 J/T is placed in uniform field of 0.15 T. If the bar is free to rotate in plane of field then which orientation would correspond to its (i) stable and (ii) unstable equilibrium? What is potential energy of magnet in each case?**

**Sol. **(i) If M is parallel to B then $\theta=0^{\circ}$ So potential energy $U=U_{\min }=-M B$

$U_{\min }=-M B=-0.32 \times 0.15 J=-4.8 \times 10^{-2} J$

This is case of stable equilibrium

(ii) If M is antiparallel to B then $\theta=\pi^{\circ}$ so potential energy

$U=U_{\max }=+M B=+0.32 \times 0.15=4.8 \times 10^{-2} J$

This is case of unstable equilibrium.

**Also Read:**

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

About eSaral

At 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.

For free video lectures and complete study material, Download** eSaral APP.**