- 5 cm
- 20 cm
- 15 cm
- 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 $