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QJEEM24Phf09d: JEE Main 2024 Physics - Class 12 JEEM24Phf09d Solution: The magnetic potential due to a magnetic dipole at a point on…

The magnetic potential due to a magnetic dipole at a point on its axis situated at a distance of 20 cm20 \mathrm{~cm} from its center is 1.5×10−5 T m1.5 \times 10^{-5} \mathrm{~T} \mathrm{~m} . The magnetic moment of the dipole is _________ A m2A \mathrm{~m}^2 . (Given : μo4π=10−7TmA−1\frac{\mu_o}{4 \pi}=10^{-7} \mathrm{Tm} A^{-1} )

Step-by-Step Verified Solution

amp;V=μ04πMr2amp;⇒1.5×10−5=10−7×M(20×10−2)2amp;⇒M=1.5×10−5×20×20×10−410−7amp;M=1.5×4=6\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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