NEET-XII-Physics
36: Permanent Magnets
- #19The magnetometer of the previous problem is used with the same magnet in Tan-B position. Where should the magnet be placed to produce a 37° deflection of the needle?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\,}``
= 3.75 × 103 A-m2/T
Deflection in the magnetometer θ = 37°
From the magnetometer theory in Tan-B position, we have
`` \frac{M}{{B}_{H}}=\frac{4\,\mathrm{\,\pi \,}}{{\,\mathrm{\,\mu \,}}_{0}}{\left({d}^{2}+{l}^{2}\right)}^{3/2}\,\mathrm{\,tan\,}\theta ``
Since for the short magnet l << d, we can neglect l w.r.t. d.
Now,
`` \frac{M}{{B}_{H}}=\frac{4\,\mathrm{\,\pi \,}}{{\,\mathrm{\,\mu \,}}_{0}}{d}^{3}\,\mathrm{\,tan\,}\theta ``
`` \Rightarrow 3.75\times {10}^{3}=\frac{1}{{10}^{-7}}\times {d}^{\mathit{3}}\times 0.75``
`` \Rightarrow {d}^{3}=\frac{3.75\times {10}^{3}\times {10}^{-7}}{0.75}=5\times {10}^{-4}``
`` \Rightarrow d=\sqrt[3]{5\times {10}^{-4}}``
`` \Rightarrow d=0.079\,\mathrm{\,m\,}=7.9\,\mathrm{\,cm\,}``
Magnet will be at 7.9 cm from the centre.
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