Magnetic Effects & Magnetism

JEE Main level, numerical answer, medium

264 of 276

Question 264 of 276

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})

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Answer: 6

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}} \Rightarrow M=\dfrac{1.5\times {10}^{-5}\times 20\times 20\times {10}^{-4}}{{10}^{-7}} \Rightarrow M=1.5\times 4=6A{m}^{2}

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