CBSE Class 12-science Answered

Dear Student

A simple electric dipole in an external magnetic field performs harmonic motion (for small angles between E and L) due to the moments by the forces acting on the respective charges. Assuming there is no system to dissipate energy that is. The mechanic moment is given by p x E and the energy of the system is given by -p.E
We were asked to calculate the period of this harmonic motion.

At theta Pi/2, both the KE and the energy of the dipole rotation are zero, which means that their changes are equal.
When looking at theta = 0 this gives:
p*E = I * omega^2 / 2 which gives us omega after one fourth of the period.
Alpha = d omega / dt and I * Alpha = p x E .
So Alpha * (t2-t1) = omega2-omega1 (approximately) and Alpha = p x E /I
Putting these together and using omega1 = 0:
omega2/(T/4) = sqrt(p*E*2/I)/(T/4) = p x E / I which gives:
Proceed in this way we will up date you later.

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