NEET-XII-Physics
36: Permanent Magnets
- #20A deflection magnetometer is placed with its arms in north-south direction. How and where should a short magnet having M/BH = 40 A m2 T-1 be placed so that the needle can stay in any position?Ans : Given:
Ratio, `` \frac{M}{{B}_{H}}=40{\,\mathrm{\,Am\,}}^{2}/\,\mathrm{\,T\,}``
Since the magnet is short, l can be neglected.
So, using the formula for `` \frac{\mathit{M}}{{\mathit{B}}_{\mathit{H}}}`` from the magnetometer theory and substituting all values, we get
`` \frac{M}{{B}_{H}}=\frac{4\,\mathrm{\,\pi \,}}{{\,\mathrm{\,\mu \,}}_{0}}\frac{{d}^{3}}{2}=40``
`` \Rightarrow {d}^{3}=40\times {10}^{-7}\times 2``
`` \Rightarrow {d}^{3}=8\times {10}^{-6}``
`` \Rightarrow d=2\times {10}^{-2}\,\mathrm{\,m\,}=2\,\mathrm{\,cm\,}``
Thus, the magnet should be placed in such a way that its north pole points towards the south and it is 2 cm away from the needle.
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