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Physics Question on Magnetism and matter

A short bar magnet has a magnetic moment of 0.48 J T−10.48\, J\, T^{-1}. The magnitude and direction of magnetic field produced by the magnet at a distance of 10 cm10 \,cm from the centre of the magnet on its axis is

A

0.48×10−4 T0.48 \times 10^{-4}\, T along N−SN-S direction

B

0.28×10−4 T0.28 \times 10^{-4}\, T along S−NS-N direction

C

0.28×10−4 T0.28 \times 10^{-4}\, T along N−SN-S direction

D

0.96×10−4 T0.96 \times 10^{-4}\, T along S−NS-N direction

Answer

0.96×10−4 T0.96 \times 10^{-4}\, T along S−NS-N direction

Explanation

Solution

On the axis of the magnet B=μ04π⋅2md3B = \frac{\mu_{0}}{4\pi} \cdot \frac{2m}{d^{3}} Here, μ04π=10−7 A m−2\frac{\mu _{0}}{4\pi } = 10^{-7} \, A\,m^{-2} m=0.48 J T−1,d=10 cm=0.1 mm = 0.48 \,J \,T^{-1}, d = 10 \,cm = 0.1 \,m Then, B=10−7×2×0.48(0.1)3B = \frac{10^{-7}\times 2 \times 0.48 }{\left(0.1\right)^{3}} =0.96×10−4 T= 0.96 \times 10^{-4}\,T, along S−NS-N direction.