A bar magnet of magnetic moment At and moment M of inertia I is freely suspended such that the magnetic axial line is in the direction of magnetic meridian. If the magnet is displaced by a very small angle θ , angular acceleration is (magnetic induction of earth's horizontal field =BH)

Question

A bar magnet of magnetic moment At and moment M of inertia I is freely suspended such that the magnetic axial line is in the direction of magnetic meridian. If the magnet is displaced by a very small angle θ , angular acceleration is (magnetic induction of earth's horizontal field =BH) (A)  (MBHθ /I)  (B)  (IBHθ /M)  (C)  (Mθ /IBH)  (D)  (Iθ /MBH)

Tardigrade Admin · Accepted Answer

Restoring torque on the magnet, τ=-2 m BH l sin θ=-M BH sin θ if θ is small sin θ ≈ θ I α= -M BH θ ; α=(M BH θ/I)

$\frac{M B _{H} θ}{I}$

$\frac{I B _{H} θ}{M}$

$\frac{Mθ}{I B _{H}}$

$\frac{I θ}{M B _{H}}$

Restoring torque on the magnet,
$τ = - 2 m B_{H} l sin θ = - M B_{H} sin θ$
if $θ$ is small $sin θ \approx θ$
$I α = - M B_{H} θ; α = \frac{M B _{H} θ}{I}$

Q. A bar magnet of magnetic moment At and moment M of inertia I is freely suspended such that the magnetic axial line is in the direction of magnetic meridian. If the magnet is displaced by a very small angle θ, angular acceleration is (magnetic induction of earth's horizontal field =BH​)

IMBH​θ​

MIBH​θ​

IBH​Mθ​

MBH​Iθ​

Restoring torque on the magnet, τ=−2mBH​lsinθ=−MBH​sinθ if θ is small sinθ≈θ Iα=−MBH​θ;α=IMBH​θ​

$\frac{M B _{H} θ}{I}$

$\frac{I B _{H} θ}{M}$

$\frac{Mθ}{I B _{H}}$

$\frac{I θ}{M B _{H}}$

Restoring torque on the magnet,
$τ = - 2 m B_{H} l sin θ = - M B_{H} sin θ$
if $θ$ is small $sin θ \approx θ$
$I α = - M B_{H} θ; α = \frac{M B _{H} θ}{I}$