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PhysicsMagnetismMagnetic Dipole & MomentMedium2 minPYQ_2020

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If at a distance of $40 cm$ at an axial position of a dipole, the "magnetic potential" (analogous to electric potential) is $2 .4 \times 10^{- 5} J A m^{- 1}$ , then the magnetic moment of the dipole is

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Answer:

Solution:

Here, $r = 40 c m = 0.4 m$
$θ = 0 °$ (an axial line)
$M = ?$

As $V = \frac{μ_{0}}{4 π} \frac{M c o s θ}{r^{2}}$
$\Rightarrow 2.4 \times 10^{- 5} = 10^{- 7} \times \frac{M \times 1}{{(0.4)}^{2}}$
$\Rightarrow M = 38.4 A m^{2}$

Stream:JEESubject:PhysicsTopic:MagnetismSubtopic:Magnetic Dipole & Moment

⏱ 2mℹ️ Source: PYQ_2020

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