NEET-XII-Physics
36: Permanent Magnets
- #9A magnetic dipole of magnetic moment 0.72 A m2 is placed horizontally with the north pole pointing towards south. Find the position of the neutral point if the horizontal component of the earth’s magnetic field is 18 μT.Ans : Given:
Magnetic moment of the magnetic dipole, M = 0.72 Am2
Horizontal component of Earth's magnetic field, BH = 18 μT
Let d be the distance of the neutral point from the south of the dipole.
When the magnet is such that its north pole faces the geographic south of Earth, the neutral point lies along the axial line of the magnet.
Thus, the magnetic field on the axial point of the dipole (B) is given by
`` B=\frac{{\,\mathrm{\,\mu \,}}_{0}}{4\,\mathrm{\,\pi \,}}\frac{2M}{{d}^{3}}``
`` ``
This magnetic field strength should be equal to the horizontal component of Earth's magnetic field.
Thus,
`` \frac{{10}^{-7}\times 2\times 0.72}{{d}^{3}}=18\times {10}^{-6}``
`` \Rightarrow {d}^{3}=\frac{2\times 0.72\times {10}^{-7}}{18\times {10}^{-6}}``
`` \Rightarrow d={\left(\frac{8\times {10}^{-9}}{{10}^{-6}}\right)}^{1/3}``
`` \Rightarrow d=2\times {10}^{-1}\,\mathrm{\,m\,}=20\,\mathrm{\,cm\,}``
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