A short bar magnet has a magnetic moment of $0.48$ J $T^{- 1}$ . The magnitude and direction of magnetic field produced by the magnet at a distance of $10$ cm from the centre of the magnet at a magnet on its axis is?

0.48 \times 10^{- 4}

T along N-S direction

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0.28 \times 10^{- 4}

T along S-N direction

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0.28 \times 10^{- 4}

T along N-S direction

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0.96 \times 10^{- 4}

T along S-N direction

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Solution

The correct option is D $0.96 \times 10^{- 4}$ T along S-N direction
On the axis of the magnet
$B = \frac{μ_{o}}{4 π} \cdot \frac{2 m}{d^{3}}$
Here, $\frac{μ_{o}}{4 π} = 10^{- 7}$ A $m^{- 2}$
$m = 0.48$ J $T^{- 1}$ , $d = 10$ cm $= 0.1$ m
Then, $B = \frac{10^{- 7} \times 2 \times 0.48}{(0.1)^{3}}$
$= 0.96 \times 10^{- 4}$ T, along S-N direction.

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