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5.Magnetism and Matter
hard
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 $1.5 \times 10^{-5} \ \mathrm{Tm}$. The magnetic moment of the dipole is___________ $\mathrm{Am}^2$. (Given : $\frac{\mu_0}{4 \pi}=10^{-7} \ \mathrm{TmA}^{-1}$ )
A
$6$
B
$5$
C
$4$
D
$12$
(JEE MAIN-2024)
Solution
$ \mathrm{V}=\frac{\mu_0}{4 \pi} \frac{\mathrm{M}}{\mathrm{r}^2} $
$ \Rightarrow 1.5 \times 10^{-5}=10^{-7} \times \frac{\mathrm{M}}{\left(20 \times 10^{-2}\right)^2} $
$ \Rightarrow \mathrm{M}=\frac{1.5 \times 10^{-5} \times 20 \times 20 \times 10^{-4}}{10^{-7}} $
$ \mathrm{M}=1.5 \times 4=6$
Standard 12
Physics
