Question
The magnetometer of the previous problem is used with the same magnet in $\tan-\text{B}$ position. Where should the magnet be placed to produce a 37° deflection of the needle?
✓
Answer
Given $\frac{\text{M}}{\text{B}_\text{H}}=40\text{A-m}^2/\text{T}$
Since the magnet is short ‘ℓ’ can be neglected
So, $\frac{\text{M}}{\text{B}_\text{H}}=\frac{4\pi}{\mu_0}\times\frac{\text{d}^3}{2}=40$
$\Rightarrow\text{d}^3=\frac{40\times4\pi\times10^{-7}\times2}{4\pi}=8\times10^{-6}$
$\Rightarrow\text{d}=2\times10^{-2}\text{m}=2\text{cm}$
with the northpole pointing towards south.
Since the magnet is short ‘ℓ’ can be neglected
So, $\frac{\text{M}}{\text{B}_\text{H}}=\frac{4\pi}{\mu_0}\times\frac{\text{d}^3}{2}=40$
$\Rightarrow\text{d}^3=\frac{40\times4\pi\times10^{-7}\times2}{4\pi}=8\times10^{-6}$
$\Rightarrow\text{d}=2\times10^{-2}\text{m}=2\text{cm}$
with the northpole pointing towards south.
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