A deflection magnetometer is placed with its arms in north-south direction. How and where should a short magnet having $\frac{M}{B_{H}} = 40 A m^{2} T^{- 1}$ be placed so that the needle can stay in any position ?

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Solution

Given, $\frac{M}{B_{H}} = 40 \frac{A - m^{2}}{T}$

Since the magnet is short l can be neglected

So, $\frac{M}{B_{H}} = \frac{4 π}{μ_{0}}, \frac{d^{3}}{2} = 40$

$\Rightarrow d^{3} = \frac{40 \times 4 π \times 10^{- 7} \times 2}{0.75}$

= $8 \times 10^{- 6}$

d= $2 \times 10^{- 2} m$ = 2 cm

with the north pole pointing towards south.

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A deflection magnetometer is placed with its arms in north-south direction. How and where should a short magnet having MBH=40Am2T−1 be placed so that the needle can stay in any position ?

Given, MBH=40A−m2T Since the magnet is short l can be neglected So, MBH=4πμ0,d32=40 ⇒d3=40×4π×10−7×20.75 = 8×10−6 d= 2×10−2m = 2 cm with the north pole pointing towards south.

A deflection magnetometer is placed with its arms in north-south direction. How and where should a short magnet having $\frac{M}{B_{H}} = 40 A m^{2} T^{- 1}$ be placed so that the needle can stay in any position ?