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JEE Main 202429 JanuaryMorning Shift
Exam
JEE Main
Year
2024
Date
29 January
Shift
Morning Shift
Subject
Physics
Chapter
Magnetic Effects of Current

JEE Main 2024 · 29 January · Morning Shift · Physics · Magnetic Effects of Current

The magnetic potential due to a magnetic dipole at a point on its...

The magnetic potential due to a magnetic dipole at a point on its axis situated at a distance of 20cm20cm from its center is 1.5×10−5Tm1 . 5 \times {10}^{- 5}Tm. The magnetic moment of the dipole is _______ Am2A{m}^{2}. (Given : μ04π=10−7TmA−1\dfrac{{\mu}_{0}}{4\pi} = {10}^{- 7}Tm{A}^{- 1})

Correct answer

0

Step-by-step solution

The formula to calculate the magnetic potential due to a magnetic dipole can be written as

V=μ04πMr2...(1)V = \dfrac{{\mu}_{0}}{4\pi}\dfrac{M}{{r}^{2}} . . . \left(1\right)

Substitute the values of the known parameters into equation (1) and solve to calculate the required magnetic moment of the dipole.

1.5×10−5=10−7×M(20×10−2)2⇒M=1.5×10−5×20×20×10−410−7⇒M=1.5×4=6Am21 . 5 \times {10}^{- 5} = {10}^{- 7} \times \dfrac{M}{{\left(20 \times {10}^{- 2}\right)}^{2}} ⇒ M = \dfrac{1 . 5 \times {10}^{- 5} \times 20 \times 20 \times {10}^{- 4}}{{10}^{- 7}} ⇒ M = 1 . 5 \times 4 = 6A{m}^{2}

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