NEET-XII-Physics
36: Permanent Magnets
- #10A magnetic dipole of magnetic moment
0.722 A m2is placed horizontally with the north pole pointing towards east. 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 magnetic dipole, M `` =0.72\sqrt{2}\,\mathrm{\,A\,}-{\,\mathrm{\,m\,}}^{2}``
Horizontal component of the earth's magnetic field, BH = 18 μT
Let d be the distance of neutral point from the dipole.
Magnetic field due to the bar magnet `` \left(B\right)`` on the equatorial line of the dipole is given by,
`` B=\frac{{\mu }_{0}}{4\pi }\frac{M}{{d}^{3}}``
`` \Rightarrow \frac{4\,\mathrm{\,\pi \,}\times {10}^{-7}}{4\,\mathrm{\,\pi \,}}\times \frac{0.72\sqrt{2}}{{d}^{3}}=18\times {10}^{-6}``
`` \Rightarrow {d}^{3}=\frac{0.72\times 1.414\times {10}^{-7}}{18\times {10}^{-6}}``
`` \Rightarrow {d}^{3}=0.005656``
`` \Rightarrow d=\sqrt[3]{0.005656}``
`` \Rightarrow d\approx 0.2\,\mathrm{\,m\,}=20\,\mathrm{\,cm\,}``
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