Askiitians Tutor Team
Last Activity: 5 Months ago
To determine the angular velocity of a bar magnet as it passes through the direction of a uniform magnetic field after being released from a position at right angles to the field, we can apply principles from physics, particularly those related to torque, energy conservation, and angular motion.
Understanding the System
When the bar magnet is turned through 90 degrees and released, it experiences a torque due to the magnetic field. This torque will cause the magnet to rotate until it aligns with the magnetic field. The key parameters involved in this scenario are:
- Magnetic Moment (m): This is a measure of the strength of the magnet.
- Moment of Inertia (I): This quantifies the resistance of the magnet to angular acceleration about its center of mass.
- Magnetic Field (B): The uniform magnetic field in which the magnet is placed.
Torque Acting on the Magnet
The torque (\( \tau \)) acting on the magnet when it is at an angle \( \theta \) to the magnetic field is given by:
\( \tau = m \cdot B \cdot \sin(\theta) \)
At the moment of release, the magnet is at 90 degrees to the magnetic field, so \( \sin(90^\circ) = 1 \). Thus, the torque is maximized:
\( \tau = m \cdot B \)
Angular Motion and Energy Conservation
As the magnet rotates, it converts potential energy into kinetic energy. The potential energy (U) of the magnet when it is displaced from the equilibrium position can be expressed as:
\( U = -m \cdot B \cdot \cos(\theta) \)
When the magnet is at 90 degrees, the potential energy is at its maximum. As it rotates down to align with the magnetic field (0 degrees), this potential energy is converted into rotational kinetic energy (K):
\( K = \frac{1}{2} I \omega^2 \)
Where \( \omega \) is the angular velocity we want to find.
Setting Up the Energy Equation
At the point of release (90 degrees), the potential energy is:
\( U = -m \cdot B \cdot \cos(90^\circ) = 0 \)
As the magnet passes through the direction of the field (0 degrees), the potential energy becomes:
\( U = -m \cdot B \cdot \cos(0^\circ) = -m \cdot B \)
Thus, the change in potential energy as it moves from 90 degrees to 0 degrees is:
\( \Delta U = 0 - (-m \cdot B) = m \cdot B \)
Setting this equal to the kinetic energy gives us:
\( m \cdot B = \frac{1}{2} I \omega^2 \)
Solving for Angular Velocity
Rearranging the equation to solve for \( \omega \), we have:
\( \omega^2 = \frac{2mB}{I} \)
Taking the square root of both sides, we find:
\( \omega = \sqrt{\frac{2mB}{I}} \)
Final Thoughts
This formula gives us the angular velocity of the magnet as it passes through the direction of the magnetic field. It highlights the relationship between the magnetic moment, the magnetic field strength, and the moment of inertia, illustrating how these factors influence the motion of the magnet. Understanding these principles not only helps in solving this problem but also lays the groundwork for exploring more complex dynamics in magnetic systems.
