NEET-XII-Physics
36: Permanent Magnets
- #8A magnetic dipole of magnetic moment 1.44 A m2 is placed horizontally with the north pole pointing towards north. 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 = 1.44 A-m2
Horizontal component of Earth's magnetic field, BH = 18 μT
We know that for a magnetic dipole with pole pointing to the north, the neutral point always lies in the broadside-on position.
Let d be the perpendicular distance of the neutral point from mid point of the magnet.
The magnetic field due to the dipole at the broadside-on position (B) is given by
`` \stackrel{\mathit{\to }}{B}=\frac{{\,\mathrm{\,\mu \,}}_{0}M}{4\,\mathrm{\,\pi \,}{d}^{3}}``
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This magnetic field strength should be equal to the horizontal component of Earth's magnetic field at that point, that is, BH due to Earth.
Thus,
`` \stackrel{\mathit{\to }}{B}=\frac{{\,\mathrm{\,\mu \,}}_{0}M}{4\,\mathrm{\,\pi \,}{d}^{3}}``
`` \Rightarrow B=\frac{{10}^{-7}\times 1.44}{{d}^{3}}``
`` \Rightarrow 18\times {10}^{-6}=\frac{{10}^{-7}\times 1.44}{{d}^{3}}``
`` \Rightarrow {d}^{\mathit{3}}=8\times {10}^{-3}``
`` \Rightarrow d=2\times {10}^{-1}=20\,\mathrm{\,cm\,}``
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