JEE Main & Advanced Physics Magnetism Force and Field

(1) Coulombs law in magnetism : The force between two magnetic poles of strength m1 and m2 lying at a distance \[r\] is given by \[F=k.\frac{{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}\]. In S.I. units \[k=\frac{{{\mu }_{0}}}{4\pi }={{10}^{-7}}wb/Amp\times m\], In CGS units \[k=1\]

(2) Magnetic field

(i) Magnetic field due to an imaginary magnetic pole (Pole strength m) : Is given by \[B=\frac{F}{{{m}_{0}}}\] also\[B=\frac{{{\mu }_{0}}}{4\pi }.\frac{m}{{{d}^{2}}}\]

(ii) Magnetic field due to a bar magnet : At a distance r from the centre of magnet

(a) On axial position

\[{{B}_{a}}=\frac{{{\mu }_{0}}}{4\pi }\frac{2Mr}{{{({{r}^{2}}-{{l}^{2}})}^{2}}}\]; If \[l<<r\] then \[{{B}_{a}}=\frac{{{\mu }_{0}}}{4\pi }\frac{2M}{{{r}^{3}}}\]

(b) On equatorial position : \[{{B}_{e}}=\frac{{{\mu }_{0}}}{4\pi }\frac{M}{{{({{r}^{2}}+{{l}^{2}})}^{3/2}}}\]; If \[l<<r\,;\] then \[{{B}_{e}}=\frac{{{\mu }_{0}}}{4\pi }\frac{M}{{{r}^{3}}}\]

(c) General position :  In general position for a short bar magnet \[{{B}_{g}}=\frac{{{\mu }_{0}}}{4\pi }\frac{M}{{{r}^{3}}}\sqrt{(3{{\cos }^{2}}\theta +1)}\]

(3) Bar magnet in magnetic field : When a bar magnet is left free in an uniform magnetic field, if align it self in the directional field.

(i) Torque : \[\tau =MB\sin \theta \Rightarrow \overrightarrow{\tau }=\overrightarrow{M}\times \overrightarrow{B}\]

(ii) Work :  \[W=MB(1-\cos \theta )\]

(iii) Potential energy : \[U=-MB\cos \theta =-\overrightarrow{M}\,.\,\overrightarrow{B}\]; (\[\theta =\] Angle made by the dipole with the field)

(4) Gauss's law in magnetism : Net magnetic flux through any closed surface is always zero i.e. \[\oint{\overrightarrow{B}.\overrightarrow{ds}=0}\]