NEET Physics Magnetism and Matter Questions Solved

$$\begin{gathered} \mathrm{Initial} \mathrm{dipole} \mathrm{moment} = \mathrm{M}=\mathrm{m}.\mathrm{l} \left(\mathrm{m}=\mathrm{pole} \mathrm{strength}\right) \\ \mathrm{arc}=\mathrm{angle}\times \mathrm{radius} \\ \mathrm{l}=\frac{2\mathrm{\pi }}{3}\times \mathrm{r} \\ \mathrm{r}=\frac{3\mathrm{l}}{2\mathrm{\pi }} \\ \mathrm{Now}, \mathrm{as} \mathrm{the} \mathrm{distance} \mathrm{b}/\mathrm{w} \mathrm{poles} = \mathrm{r} \\ \mathrm{New} \mathrm{dipole} \mathrm{moment} = \mathrm{m}\times \mathrm{r} = \mathrm{m}\times \frac{3\mathrm{l}}{2\mathrm{\pi }} \\ =\mathrm{m}\times \mathrm{l}\times \frac{3}{2\mathrm{\pi }} \\ =\frac{3\mathrm{M}}{2\mathrm{\pi }}\mathrm{mol} \end{gathered}$$

(3)

$$\begin{gathered} At neutral point, magnetic field due to the bar magnet is equal to horizontal component of Earth\text{'}s magentic field. \\ Initially, when bar magnet\text{'}s north pole faces magnetic north pole of the Earth: \\ \frac{2M}{r^3}=H \\ Finally, when the bar magnet is rotated by {180}^0: \\ \frac{M}{r{\text{'}}^3}=H=\frac{2M}{r^3} \\ r\text{'}=2^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}r \end{gathered}$$

If we define magnetic moment like electric dipole moment

Magnetic moment ( Ampere X $m^2$ ) = Pole strength X Distance (m)

Unit of pole strength = Ampere X metre

(1)

$$\begin{gathered} \mathrm{At} \mathrm{axial} \mathrm{point}: \mathrm{B}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{2\mathrm{M}}{{\mathrm{r}}^3} \\ \mathrm{At} \mathrm{equitorial} \mathrm{point} (\mathrm{at} \mathrm{same} \mathrm{distance}): \mathrm{B}\text{'}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{\mathrm{M}}{{\mathrm{r}}^3}=\frac{1}{2}\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{2\mathrm{M}}{{\mathrm{r}}^3}=\frac{1}{2}\mathrm{B} \\ \mathrm{At} \mathrm{distance} 2\mathrm{r}: \\ \mathrm{B}\text{'}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{\mathrm{M}}{{\left(2\mathrm{r}\right)}^3}==\frac{1}{8}\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{\mathrm{M}}{{\mathrm{r}}^3}=\frac{1}{8}\frac{\mathrm{B}}{2}=\frac{\mathrm{B}}{16} \end{gathered}$$