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Physics · Magnetism and Matter

JEE Main 2024 — 29 January, Shift 1 — Question 53

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 1.5×10−5Tm1.5 \times 10^{-5} \mathrm{Tm}. The magnetic moment of the dipole is \qquad Am2\mathrm{Am}^{2}. (Given : μ04π=10−7TmA−1\frac{\mu_{0}}{4 \pi}=10^{-7} \mathrm{TmA}^{-1} )

Answer: 6

Numerical answer — enter this value.

Step-by-step solution

V=μ04πMr2\mathrm{V}=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{M}}{\mathrm{r}^{2}}

⇒1.5×10−5=10−7×M(20×10−2)2\Rightarrow 1.5 \times 10^{-5}=10^{-7} \times \frac{\mathrm{M}}{\left(20 \times 10^{-2}\right)^{2}} ⇒M=1.5×10−5×20×20×10−410−7\Rightarrow \mathrm{M}=\frac{1.5 \times 10^{-5} \times 20 \times 20 \times 10^{-4}}{10^{-7}} M=1.5×4=6\mathrm{M}=1.5 \times 4=6

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
29 January, Shift 1
Subject
Physics
Chapter
Magnetism and Matter
Topic
Natural Magnetism: Bar Magnets