Question

Obtain the expression for the period of a magnet vibrating in a uniform magnetic field and performing S.H.M.

Answer 1

The expression is given as T = `2pi sqrt("I"/"μB")`

Explanation:

The time period of oscillation of a magnet in a uniform magnetic field 'B' is given by

Formula:

T = `2pi sqrt("I"/"μB")`

Where

• T = Time period
• I = Moment of inertia
• μ = magnetic dipole moment
• B = magnetic field 

The time period of an oscillating body about a fixed point can be defined as the time taken by the body to complete one vibration around that particular point, called the time period.

Answer 2

1. If a bar magnet is freely suspended in the plane of a uniform magnetic field, it remains in equilibrium with its axis parallel to the direction of the field.
   If it is given a small angular displacement θ about an axis passing through its centre, perpendicular to itself and to the field, and released, it performs angular oscillations.
2. Let μ be the magnetic dipole moment and B the magnetic field. In the deflected position, a restoring torque acts on the magnet that tends to bring it back to its equilibrium position.
   
   Magnet vibrating in a uniform magnetic field
3. The magnitude of this torque is τ = µBsinθ
   If θ is small, sinθ ≈ θ
   ∴ τ = µBθ
4. For clockwise angular displacement θ, the restoring torque is in the anticlockwise direction.
   ∴ τ = Iα = –μBθ
   Where I is the moment of inertia of the bar magnet and α is its angular acceleration.
   ∴ α = `-((mu"B")/"I")θ` ....(1)
5. Since µ, B, and I are constants, equation (1) shows that angular acceleration is directly proportional to the angular displacement and directed opposite to the angular displacement. Hence the magnet performs angular S.H.M.
6. The period of vibrations of the magnet is given by
   T = `(2pi)/sqrt("angular acceleration per unit angular displacement")`
   = `(2pi)/sqrt(alpha"/"theta)`
   Thus, by considering the magnitude of angular acceleration (α),
   T = `2pisqrt(("I")/(mu"B"))`