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

10 cm

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20 cm

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30 cm

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40 cm

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Solution

The correct option is C $30 cm$
At neutral point,

$| B_{H} | = | B_{m a g n e t} |$

Let $d$ be the distance of the neutral point from the magnet on it axis.

Magnetic field due to a magnet on an axial point is given by,

$B_{m a g n e t} = \frac{μ_{0}}{4 π} \frac{2 M}{d^{3}}$

$\Rightarrow 10^{- 7} \times \frac{2 \times 6.75}{d^{3}} = 5 \times 10^{- 5}$

$d = 0.3 m = 30 cm$

Hence, option $(C)$ is the correct answer.

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A short bar magnet of moment 6.75 A-m2, produces a neutral point on its axis. If the horizontal component of earth's magnetic field is 5×10−5 Wb/m2, then the distance of the neutral point should be

The correct option is C 30 cmAt neutral point, |BH|=|Bmagnet| Let d be the distance of the neutral point from the magnet on it axis. Magnetic field due to a magnet on an axial point is given by, Bmagnet=μ04π2Md3 ⇒10−7×2×6.75d3=5×10−5 d=0.3 m=30 cm Hence, option (C) is the correct answer.

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