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 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
$B = \frac{μ_{0}}{4 π} . \frac{2 m}{d^{3}}$
Here, $\frac{μ_{0}}{4 π} = 10^{- 7} A m^{- 2}$
$m = 0.48 J T^{- 1}, d = 10 c m = 0.1 m$
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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Similar questions

Q. A short bar magnet has a magnetic moment of

0.48

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?

A bar magnet with magnetic moment M is placed in a space where a magnetic field $B_{1}$ is acting in $30^{0}$ anticlockwise to S-N direction and another magnetic field $B_{2}$ is along $45^{0}$ clockwise to S-N direction. The magnitude of $B_{1}$ is $\sqrt{2} \times 10^{- 2} T$ . What should be the magnitude of $B_{2}$ for the equilibrium of the magnet?