NEET-XII-Physics
36: Permanent Magnets
- #18A short magnet produces a deflection of 37° in a deflection magnetometer in Tan-A position when placed at a separation of 10 cm from the needle. Find the ratio of the magnetic moment of the magnet to the earth’s horizontal magnetic field.Ans : Given:
Deflection in the magnetometer in the given position when placed in the magnetic field of a short magnet, θ = 37°
Separation between the magnet and the needle, d = 10 cm = 0.1 m
Let M be the magnetic moment of the magnet and BH be Earth's horizontal magnetic field.
According to the magnetometer theory,
`` \frac{M}{{B}_{H}}=\frac{4\,\mathrm{\,\pi \,}}{{\,\mathrm{\,\mu \,}}_{0}}\frac{{\left({d}^{2}-{l}^{2}\right)}^{2}}{2d}\,\mathrm{\,tan\,}\theta ``
`` \,\mathrm{\,For\,}\,\mathrm{\,the\,}\,\mathrm{\,short\,}\,\mathrm{\,magnet\,},``
`` \frac{M}{{B}_{H}}=\frac{4\,\mathrm{\,\pi \,}}{{\,\mathrm{\,\mu \,}}_{0}}\times \frac{{d}^{4}}{2d}\,\mathrm{\,tan\,}\theta ``
`` \Rightarrow \frac{M}{{B}_{H}}=\frac{4\,\mathrm{\,\pi \,}}{4\,\mathrm{\,\pi \,}\times {10}^{-7}}\times \frac{{\left(0.1\right)}^{3}}{2}\times \,\mathrm{\,tan\,}37°``
`` \Rightarrow \frac{M}{{B}_{H}}=0.5\times 0.75\times 1\times {10}^{4}``
`` \Rightarrow \frac{M}{{B}_{H}}=3.75\times {10}^{3}\,\mathrm{\,A\,}-{\,\mathrm{\,m\,}}^{2}/\,\mathrm{\,T\,}``
Page No 278: