Question

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

Answer 1

Let a bar magnet is placed in a uniform magnetic field.

• Magnetic dipole moment of bar magnet = M
• Intensity of magnetic field = B

In a uniform magnetic field, a magnetic dipole experiences no net force but experiences a non-zero torque which is given by

$\dag\:\underline{\boxed{\bf{\orange{\tau=\overrightarrow{M}\times\overrightarrow{B}=MB\:sin\theta}}}}$

Where θ is the angle between M and B.

In SHM angle θ is small. (sin θ ≈ θ)

Therefore τ = M B θ

• Torque is measured as the product of moment of inertia and angular acceleration.

➠ I α = M B θ

➠ α = M B θ / I

Comparing with equation of angular SHM

$:\implies\:\underline{\boxed{\bf{\red{\dfrac{d^2\theta}{dt^2}+\omega^2\theta=0}}}}$

➙ ω² θ = M B θ / I

➙ ω² = M B / I

➙ ω = √M B / I

♦ Time period of angular SHM :

$\sf:\implies\:T=\dfrac{2\pi}{\omega}$

$:\implies\:\underline{\boxed{\bf{\gray{T=2\pi\sqrt{\dfrac{I}{M\:B}}}}}}$

Answer 2

Answer:

$\small{\colorbox{cyan}{✓Verified Answer}}$

AnsWer is in attachment