The time period of oscillation of a freely su | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) freely suspended bar magnet in a uniform magnetic field $B_H$ experiences a restoring torque $	au = -M B_H \sin 	heta$. For small oscillations,$

Vedclass · Accepted Answer

$T = 2\pi \sqrt{\frac{I}{M B_H}}$

Vedclass · Answer

(A) freely suspended bar magnet in a uniform magnetic field $B_H$ experiences a restoring torque $	au = -M B_H \sin 	heta$. For small oscillations,$\sin 	heta \approx 	heta$,so $	au = -M B_H 	heta$. The equation of motion is $I \alpha = 	au$,where $I$ is the moment of inertia and $\alpha$ is the angular acceleration. $I \frac{d^2 	heta}{dt^2} = -M B_H 	heta$. This is the equation of simple harmonic motion $\frac{d^2 	heta}{dt^2} + \omega^2 	heta = 0$,where $\omega^2 = \frac{M B_H}{I}$. The time period $T$ is given by $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{I}{M B_H}}$.

The time period of oscillation of a freely suspended bar magnet with usual notations is given by

Similar Questions

When two magnetic moments are compared using the equal distance method,the deflections produced are $45^\circ$ and $30^\circ$. If the lengths of the magnets are in the ratio $1:2$,the ratio of their pole strengths is:

The periodic time of a magnet in a vibration magnetometer is $T_0$. If the magnet is replaced by another magnet whose moment of inertia is three times and magnetic moment is one-third of the original magnet, then the time period of the new magnet will be:

$A$ vibration magnetometer is placed at the south pole; then its time period will be

Time period of a freely suspended magnet does not depend upon

Vedclass Test Series

Exam Paper Generator

Online Exam Module