Question

A short bar magnet has a magnetic moment of $0.65\, J \,T^{-1}$, then the magnitude and direction of the magnetic field produced by the magnet at a distance $8 \,cm$ from the centre of magnet on the axis is

• A. $2.5 \times 10^{-4} \, T$, along $NS$ direction
• B. $2.5 \times 10^{-4} \, T$, along $SN$ direction
• C. $4.5 \times 10^{-4} \, T$, along $NS$ direction
• D. $4.5 \times 10^{-4} \, T$, along $SN$ direction

Answer 1

Answer: (B)

Here, $M=0.65\, J\, T^{-1}$,
$d=8\, cm $
$=0.08\, m$
The field produced by magnet at axial point is given by
$B=\frac{\mu_{0}}{4\pi} \frac{2M}{d^{3}}=10^{-7}\times\frac{2\times0.65}{\left(0.08\right)^{3}}$
$=2.5 \times 10^{-4}\, T$ along $SN$