To find the period of small oscillations of a square current-carrying loop in a magnetic field, we need to analyze the forces and torques acting on the loop. The loop will experience a torque due to the magnetic field, which will cause it to oscillate about its stable equilibrium position. Let's break down the problem step by step.
Understanding the System
We have a square loop of wire with a mass of 10 grams (0.01 kg) and a current of 2 A flowing through it. The loop is placed in a uniform magnetic field of strength B = 0.1 T, which is perpendicular to the plane of the loop. The loop can rotate freely about a vertical axis.
Torque on the Loop
The magnetic torque (\( \tau \)) acting on the loop can be calculated using the formula:
- \( \tau = n \cdot I \cdot A \cdot B \cdot \sin(\theta) \)
Where:
- \( n \) = number of turns (1 for a single loop)
- \( I \) = current (2 A)
- \( A \) = area of the loop
- \( B \) = magnetic field strength (0.1 T)
- \( \theta \) = angle between the normal to the loop and the magnetic field
Calculating the Area
Assuming the side length of the square loop is \( a \), the area \( A \) is given by:
Finding the Moment of Inertia
The moment of inertia (\( I \)) of a square loop about its center is given by:
- \( I = \frac{1}{6} m a^2 \)
Here, \( m = 0.01 \) kg is the mass of the loop.
Equilibrium Position
At the stable equilibrium position, the torque is balanced by the gravitational torque. The gravitational force acts at the center of mass of the loop, and we can express the gravitational torque as:
- \( \tau_g = m g \cdot \frac{a}{2} \cdot \sin(\theta) \)
Where \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).
Small Angle Approximation
For small oscillations, we can use the small angle approximation, where \( \sin(\theta) \approx \theta \). Thus, the torque equations become:
- \( \tau = n \cdot I \cdot A \cdot B \cdot \theta \)
- \( \tau_g = m g \cdot \frac{a}{2} \cdot \theta \)
Setting Up the Equation of Motion
Setting the torques equal gives us:
- \( n \cdot I \cdot A \cdot B \cdot \theta = m g \cdot \frac{a}{2} \cdot \theta \)
We can simplify this to find the angular frequency \( \omega \):
- \( \omega^2 = \frac{n \cdot I \cdot A \cdot B}{m g \cdot \frac{a}{2}} \)
Finding the Period of Oscillation
The period \( T \) of oscillation is related to the angular frequency by:
- \( T = 2\pi \sqrt{\frac{I}{\tau}} \)
Substituting the values we have, we can calculate the period of small oscillations. However, we need to know the side length \( a \) of the loop to proceed with the numerical calculation. Assuming a side length of 0.1 m (10 cm) for the loop:
Final Calculation
Let's calculate the area:
- \( A = (0.1)^2 = 0.01 \, \text{m}^2 \)
Now, substituting into the equation for \( \omega^2 \):
- \( \omega^2 = \frac{1 \cdot 2 \cdot 0.01 \cdot 0.1}{0.01 \cdot 9.81 \cdot 0.05} \approx 0.0408 \, \text{s}^{-2} \)
Thus, the angular frequency \( \omega \) is:
- \( \omega \approx 0.202 \, \text{rad/s} \)
Finally, the period \( T \) is:
- \( T = 2\pi \cdot \frac{1}{\omega} \approx 31.1 \, \text{s} \)
In summary, the period of small oscillations of the square current-carrying loop in the magnetic field is approximately 31.1 seconds. This analysis shows how the interplay between magnetic forces and gravitational forces leads to oscillatory motion in the system.
