A bar magnet of magnetic moment M and moment 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), the angular acceleration is (Magnetic induction of earth's horizontal field =BH )

Question

A bar magnet of magnetic moment M and moment 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), the angular acceleration is (Magnetic induction of earth's horizontal field =BH ) (A) (M BH θ/I) (B) (I BH θ/M) (C) (M θ/I BH) (D) (I θ/M BH)

Tardigrade Admin · Accepted Answer

When magnet is displaced by a very small angle

θ

, then

r \in

storing couple acting on the magnet is

τ = - M B_{H} sin θ

Negative sign shows the restoring nature of torque.
Now since

τ = I α

and

sin θ \approx θ

for small angular displacement

∴ I α = M B_{H} θ

or

α =

angular acceleration

= \frac{M B _{H} θ}{I}

$\frac{M B _{H} θ}{I}$

$\frac{I B _{H} θ}{M}$

$\frac{Mθ}{I B _{H}}$

$\frac{I θ}{M B _{H}}$

Q. A bar magnet of magnetic moment M and moment 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), the angular acceleration is (Magnetic induction of earth's horizontal field =BH​ )

IMBH​θ​

MIBH​θ​

IBH​Mθ​

MBH​Iθ​

When magnet is displaced by a very small angle θ, then r∈ storing couple acting on the magnet is τ=−MBH​sinθ Negative sign shows the restoring nature of torque. Now since τ=Iα and sinθ≈θ for small angular displacement ∴Iα=MBH​θ or α= angular acceleration =IMBH​θ​

$\frac{M B _{H} θ}{I}$

$\frac{I B _{H} θ}{M}$

$\frac{Mθ}{I B _{H}}$

$\frac{I θ}{M B _{H}}$