A magnetic dipole with a dipole moment of magnitude 0.0192 J/T is released from rest in a uniform magnetic field of magnitude 57.1 mT. The rotation of the dipole due to the magnetic force on it is unimpeded. When the dipole rotates through the orientation where its dipole moment is aligned with the magnetic field, its kinetic energy is 0.407 mJ. (a) What is the initial angle between the dipole moment and the magnetic field? (b) What is the angle when the dipole is next (momentarily) at rest?

Question

contestada

Respuesta :

rejkjavik rejkjavik · Answer 1 · 2019-08-21T05:32:27+02:00

Answer:

A) [tex]\theta = 50.55 degree[/tex]

B) At the angle -50.55 degree K.E come to zero again

Explanation:

we know that gain in kinetic energy is given as

[tex]\Delta K = - \muBcos\theta - (-\mu Bcos \theta)[/tex][tex]= \mu B(1- cos\theta)[/tex]

therefore

[tex]cos\theta = 1 - \frac{\Delta k}{\mu B}[/tex]

[tex]= 1 - \frac{0.4*10^{-3}}{0.0192*57.1*10^{-3}}[/tex]

[tex]cos\theta = 0.635[/tex]

[tex]\theta = 50.55 degree[/tex]

b) At the angle -50.55 degree K.E come to zero again.