Question

Show that the time period (T) of oscillations of a freely suspended magnetic dipole of magnetic moment (m) in a uniform magnetic field (B) is given by `"T" = 2pi sqrt("I"/"mB")`, where I is a moment of inertia of the magnetic dipole.

Answer 1

A magnetic dipole of dipole moment m is freely suspended in a uniform magnetic field B. Let the angle between magnetic moment m and magnetic field B is theta, then torque on the dipole is,

r = m × B = mB sinθ ...(1)

Also, torque, r = Iα

` ∵alpha = ("d"^2theta)/("dt"^2)`

`∵ "r" = "I" ("d"^2theta)/("dt"^2)` .....(2)

From eq. (1) and (2), we can write,

`"I" ("d"^2theta)/("dt"^2) = -"mB"sintheta`.....(Because angular acceleration and angle θ are in opposite direction)

For very small deflections,  sinθ∼θ

Hence,

⇒ ` "I" ("d"^2theta)/("dt"^2) = -"mB"theta`

⇒ `("d"^2theta)/("dt"^2) = - ("mB"/"I")theta`

By comparing with the standard equation of SHM, we can find that,

`omega^2 = "mB"/"I"`

⇒ `omega = sqrt("mB"/"I")`

⇒ `"T" = (2pi)/omega = 2pisqrt("I"/"mB")`

⇒ `"T" = 2pisqrt("I"/"mB")`