JEE MAIN - Physics (2024 - 29th January Morning Shift - No. 26)
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 : $$\frac{\mu_o}{4 \pi}=10^{-7} \mathrm{Tm} A^{-1}$$ )
Answer
6
Explanation
$$\begin{aligned}
& \mathrm{V}=\frac{\mu_0}{4 \pi} \frac{\mathrm{M}}{\mathrm{r}^2} \\
& \Rightarrow 1.5 \times 10^{-5}=10^{-7} \times \frac{\mathrm{M}}{\left(20 \times 10^{-2}\right)^2} \\
& \Rightarrow \mathrm{M}=\frac{1.5 \times 10^{-5} \times 20 \times 20 \times 10^{-4}}{10^{-7}} \\
& \mathrm{M}=1.5 \times 4=6
\end{aligned}$$
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