First slide

Combined field of bar magnet and Earth; neutral point

Question

A Short bar magnet is placed in the magnetic meridian of the earth with north pole pointing north. Neutral points are found at a distance of 30cm from the magnet an the east west line drawn through the middle point of the magnet. The magnetic moment of the magnet in $A m^{2}$ is $(B_{H} = 3.6 \times 10^{- 5} T)$

Moderate

A

9.72
B

4.9
C

19.4
D

14.6

Solution

When North pole of a bar magnet facing North Null points are obtained an equatorial line
here $B = B_{H}$
$B_{H} = \frac{μ_{0}}{4 π} \frac{M}{d^{3}}$
Given $B_{H} = 3.6 \times 10^{- 5} T$
$\begin{array}{l} d = 30 cm \\ 3.6 \times 10^{- 5} = \frac{10^{- 7} \times M}{{(0.3)}^{3}} \\ M = \frac{3.6 \times 10^{- 5} \times 27 \times 10^{- 3}}{10^{- 7}} = 9.72 A m^{2} \end{array}$

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