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Question: short magnet of moment 6.75 Am<sup>2</sup> produces a neutral point on its axis. If horizontal compo...

short magnet of moment 6.75 Am2 produces a neutral point on its axis. If horizontal component of earth's magnetic field is 5×10−5 Wb/m25 \times 10 ^ { - 5 } \mathrm {~Wb} / \mathrm { m } ^ { 2 }, then the distance of the neutral point should be

A

10 cm

B

20 cm

C

30 cm

D

40 cm

Answer

30 cm

Explanation

Solution

At neutral point

∣ Magnetic field due  to magnet ∣=∣ Magnetic field due  to earth ∣\left| \begin{array} { c } \text { Magnetic field due } \\ \text { to magnet } \end{array} \right| = \left| \begin{array} { c } \text { Magnetic field due } \\ \text { to earth } \end{array} \right|

μ04π⋅2Md3=5×10−5⇒10−7×2×6.75d3=5×10−5\frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { 2 M } { d ^ { 3 } } = 5 \times 10 ^ { - 5 } \Rightarrow 10 ^ { - 7 } \times \frac { 2 \times 6.75 } { d ^ { 3 } } = 5 \times 10 ^ { - 5 }

⇒d=0.3 m=30 cm\Rightarrow d = 0.3 \mathrm {~m} = 30 \mathrm {~cm}