The magnetic potential at a point on the axia | vedclass

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Question

(D) The magnetic potential $V$ at a distance $d$ from the center of a bar magnet on its axial line is given by the formula:$V = \frac{\mu_0}{4\pi} \cd

Vedclass · Accepted Answer

$\frac{V}{4}$

Vedclass · Answer

(D) The magnetic potential $V$ at a distance $d$ from the center of a bar magnet on its axial line is given by the formula:$V = \frac{\mu_0}{4\pi} \cdot \frac{M}{d^2}$From this expression,it is clear that the magnetic potential is directly proportional to the magnetic dipole moment,i.e.,$V \propto M$.Given that the initial dipole moment is $M_1 = M$ and the potential is $V_1 = V$,and the new dipole moment is $M_2 = \frac{M}{4}$,we can write:$\frac{V_1}{V_2} = \frac{M_1}{M_2}$Substituting the values:$\frac{V}{V_2} = \frac{M}{M/4} = 4$Therefore,the new magnetic potential $V_2$ is:$V_2 = \frac{V}{4}$

The magnetic potential at a point on the axial line of a bar magnet of dipole moment $M$ is $V$. What is the magnetic potential due to a bar magnet of dipole moment $\frac{M}{4}$ at the same point?

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