The magnetic potential due to a magnetic dipo | vedclass

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Question

(A) The formula for the magnetic potential $V$ due to a magnetic dipole at a point on its axis at a distance $r$ from the center is given by: $V = \fr

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$6$

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(A) The formula for the magnetic potential $V$ due to a magnetic dipole at a point on its axis at a distance $r$ from the center is given by: $V = \frac{\mu_0}{4 \pi} \frac{M}{r^2}$ Given values: $V = 1.5 	imes 10^{-5} \ T \cdot m$ $r = 20 \ cm = 0.2 \ m = 2 	imes 10^{-1} \ m$ $\frac{\mu_0}{4 \pi} = 10^{-7} \ T \cdot m \cdot A^{-1}$ Substituting these values into the formula: $1.5 	imes 10^{-5} = 10^{-7} 	imes \frac{M}{(2 	imes 10^{-1})^2}$ $1.5 	imes 10^{-5} = 10^{-7} 	imes \frac{M}{4 	imes 10^{-2}}$ $M = \frac{1.5 	imes 10^{-5} 	imes 4 	imes 10^{-2}}{10^{-7}}$ $M = \frac{6 	imes 10^{-7}}{10^{-7}} = 6 \ A \cdot m^2$ Thus,the magnetic moment of the dipole is $6 \ A \cdot m^2$.

The magnetic potential due to a magnetic dipole at a point on its axis situated at a distance of $20 \ cm$ from its center is $1.5 \times 10^{-5} \ T \cdot m$. The magnetic moment of the dipole is . . . . . . $A \cdot m^2$. (Given: $\frac{\mu_0}{4 \pi} = 10^{-7} \ T \cdot m \cdot A^{-1}$)

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