A short bar magnet has a magnetic moment of 0 | vedclass

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(A) The magnetic field $B$ on the axis of a short bar magnet at a distance $d$ from its centre is given by the formula:$B = \frac{\mu_0}{4\pi} \cdot \

Vedclass · Accepted Answer

$0.96 \, gauss$

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(A) The magnetic field $B$ on the axis of a short bar magnet at a distance $d$ from its centre is given by the formula:$B = \frac{\mu_0}{4\pi} \cdot \frac{2M}{d^3}$Given:Magnetic moment $M = 0.48 \, J \, T^{-1}$Distance $d = 10 \, cm = 0.1 \, m$$\frac{\mu_0}{4\pi} = 10^{-7} \, T \, m \, A^{-1}$Substituting the values:$B = 10^{-7} \cdot \frac{2 	imes 0.48}{(0.1)^3}$$B = 10^{-7} \cdot \frac{0.96}{0.001}$$B = 10^{-7} \cdot 960 = 9.6 	imes 10^{-5} \, T$Since $1 \, T = 10^4 \, gauss$:$B = 9.6 	imes 10^{-5} 	imes 10^4 \, gauss = 0.96 \, gauss$Thus, the correct option is $A$.

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

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