A very long magnet is placed vertically with one pole on the table. A neutral point was found at $20 c m$ from the pole. What is the pole strength if the vertical component of earth's field is $0.4 \times 10^{- 4} W b m^{- 2}$ ?

16 A - m

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8 A - m

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4 A - m

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None of these

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Solution

The correct option is A $16 A - m$
Magnetic field due to pole of strength $m$ at distance $d$ is
$B = \frac{μ_{0} m}{4 π d^{2}}$
Vertical component of earth's magnetic field = $0.4 \times 10^{- 4} W b m^{- 2}$
At the neutral point, $\frac{μ_{0} m}{4 π d^{2}} = 0.4 \times 10^{- 4}$
$⟹ m = 16 A - m$

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