The ratio of magnetic fields due to a bar magnet at the two axial points P1 and P2 which are separated from each other by 10cm is 25:2. Point P1 is situated at 10cm from the centre of the magnet.Points P1and P2 are on the same side of magnet and distance of P2 from the centre is greater than distance of P1 from the centre of magnet,then magnetic length is

1. $5$ cm
2. $20$ cm
3. $15$ cm
4. $10$ cm

Solution: The correct answer is D. 10cm

We know that the magnetic field at a distance d from the magnet’s centre on the axis is given as follows for a magnet of length 2l and magnetic moment M.

$ B=\frac{{{\mu }_{0}}2Md}{4\pi {{\left( {{d}^{2}}-{{l}^{2}} \right)}^{2}}} $

$ For~{{P}_{1}},~{{d}_{1}}=10cm~ $

$ For~{{P}_{2}},~{{d}_{2}}={{d}_{1}}+10=20cm $

$ {{B}_{1}}=\frac{{{\mu }_{0}}2M{{d}_{1}}}{4\pi {{\left( {{d}_{1}}^{2}-{{l}^{2}} \right)}^{2}}} $

$ {{B}_{2}}=\frac{{{\mu }_{0}}2M{{d}_{2}}}{4\pi {{\left( {{d}_{2}}^{2}-{{l}^{2}} \right)}^{2}}} $

$ Given:\,\,\,\frac{{{B}_{1}}}{{{B}_{2}}}=\frac{25}{2} $

$ \frac{{{d}_{1}}{{\left( {{d}_{2}}^{2}-{{l}^{2}} \right)}^{2}}}{{{d}_{2}}{{\left( {{d}_{1}}^{2}-{{l}^{2}} \right)}^{2}}}=\frac{25}{2} $

$ \frac{10{{\left( {{20}^{2}}-{{l}^{2}} \right)}^{2}}}{20{{\left( {{10}^{2}}-{{l}^{2}} \right)}^{2}}}=\frac{25}{2} $

$ \therefore l=5\,cm $

$ Hence,magnetic\text{ }length=~2l=10cm $