Question 140: Magnetism and Matter

Question

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

Accepted Answer

$$\begin{aligned}
& \mathrm{V}=\frac{\mu_0}{4 \pi} \frac{\mathrm{M}}{\mathrm{r}^2} \
& \Rightarrow 1.5 	imes 10^{-5}=10^{-7} 	imes \frac{\mathrm{M}}{\left(20 	imes 10^{-2}ight)^2} \
& \Rightarrow \mathrm{M}=\frac{1.5 	imes 10^{-5} 	imes 20 	imes 20 	imes 10^{-4}}{10^{-7}} \
& \mathrm{M}=1.5 	imes 4=6
\end{aligned}$$

Magnetism and Matter

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

Integer / Numeric Answer Type