Question
A deflection magnetometer is placed with its arms in north-south direction. How and where should a short magnet having $\frac{\text{M}}{\text{B}_\text{H}}=40\text{A-m}^2\text{T}$ be placed so that the needle can stay in any position?
✓
Answer
According to oscillation magnetometer,$\text{T}=2\pi \sqrt{\frac{\text{I}}{\text{MB}_\text{H}}}$
$\Rightarrow\frac{\pi}{10}=2\pi\sqrt{\frac{1.2\times10^{-4}}{\text{M}\times30\times10^{-6}}}$
$\Rightarrow\Big(\frac{1}{20}\Big)^2=\frac{1.2\times1^{-4}}{\text{M}\times30\times10^{-6}}$
$\Rightarrow\text{M}=\frac{1.2\times10^{-4}\times400}{30\times10^{-6}}$
$=16\times10^{2}\text{A-m}^2=1600\text{A-m}^2$
$\Rightarrow\frac{\pi}{10}=2\pi\sqrt{\frac{1.2\times10^{-4}}{\text{M}\times30\times10^{-6}}}$
$\Rightarrow\Big(\frac{1}{20}\Big)^2=\frac{1.2\times1^{-4}}{\text{M}\times30\times10^{-6}}$
$\Rightarrow\text{M}=\frac{1.2\times10^{-4}\times400}{30\times10^{-6}}$
$=16\times10^{2}\text{A-m}^2=1600\text{A-m}^2$
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