Start Free Mock LoginAlready registered? Login here

Physics - Magnetism Question with Solution | TestHub

PhysicsMagnetismMagnetic Dipole & MomentEasy2 minPYQ_2020

PhysicsEasysingle choice

The magnetic potential due to a magnetic dipole at a point on its axis distant $40 cm$ from its centre is found to be $2.4 \times 10^{- 5} J A m^{- 1}$ .The magnetic moment of the dipole will be

Options:

Answer:

Solution:

Here, $r = 40 c m = 0.4 m$

$θ = 0 °$ (an axial line)

$M = ?$

As $V = \frac{μ_{0}}{4 π} \frac{M c o s θ}{r^{2}}$

$2.4 \times 10^{- 5} = 10^{- 7} \frac{M \times 1}{{(0.4)}^{2}}$

$M = 38.4$ $A m^{2}$

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

⏱ 2mℹ️ Source: PYQ_2020

Doubts & Discussion

Loading discussions...