The magnetic field due to a short bar magnet | vedclass

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(B) The magnetic field $B$ on the axial line of a bar magnet at a distance $Z$ from its centre is given by the formula $B = \frac{\mu_0}{4 \pi} \frac{

Vedclass · Accepted Answer

$\frac{2 \mu_0}{4 \pi} \frac{M}{Z^3} \hat{M}$

Vedclass · Answer

(B) The magnetic field $B$ on the axial line of a bar magnet at a distance $Z$ from its centre is given by the formula $B = \frac{\mu_0}{4 \pi} \frac{2MZ}{(Z^2 - l^2)^2}$.For a short bar magnet,the length $2l$ is very small compared to the distance $Z$,i.e.,$Z \gg l$.Therefore,we can approximate $(Z^2 - l^2)^2 \approx (Z^2)^2 = Z^4$.Substituting this into the formula,we get $B = \frac{\mu_0}{4 \pi} \frac{2MZ}{Z^4} = \frac{\mu_0}{4 \pi} \frac{2M}{Z^3}$.In vector form,this is expressed as $\vec{B} = \frac{2 \mu_0}{4 \pi} \frac{\vec{M}}{Z^3}$.

The magnetic field due to a short bar magnet of magnetic dipole moment $M$ and length $2l$,on the axis at a distance $Z$ (where $Z \gg l$) from the centre of the magnet is given by . . . . . . .

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