Pendulum Induced EMF Calculation

This problem requires calculating the maximum induced electromotive force (EMF) in a simple pendulum moving within a uniform magnetic field.

Relevant Physics Concepts

When a conductor moves through a magnetic field, an EMF is induced across it. For a simple pendulum of length $L$ oscillating with maximum angular velocity $\omega_{max}$ in a magnetic field $B$ perpendicular to the plane of oscillation, the maximum induced EMF ($E_{max}$) is given by the formula:

$ E_{max} = \frac{1}{2} B L^2 \omega_{max} $

The maximum angular velocity ($\omega_{max}$) depends on the initial angle of release ($\theta_0$) and gravity ($g$).

Determining Maximum Angular Velocity ($\omega_{max}$)

Using conservation of energy, the potential energy lost when the pendulum swings from its highest point (angle $\theta_0$) to the lowest point (angle $0^\circ$) is converted into kinetic energy. The relationship is:

$ mgL(1 - \cos\theta_0) = \frac{1}{2} I \omega_{max}^2 $

For a simple pendulum where $I = mL^2$ (moment of inertia), this simplifies to:

$ \omega_{max}^2 = \frac{2g(1 - \cos\theta_0)}{L} $

Step-by-Step Solution

1. Gather Given Values:
   • Magnetic field strength, $B = 2 \text{ T}$
   • Pendulum wire length, $L = 10 \text{ cm} = 0.1 \text{ m}$
   • Initial angle, $\theta_0 = 60^\circ$
   • Acceleration due to gravity, $g = 10 \text{ m/s}^2$
   • Mass $m = 10 \text{ g}$ (Note: mass is not required for this calculation)
2. Calculate $(1 - \cos\theta_0)$:

   $ 1 - \cos(60^\circ) = 1 - \frac{1}{2} = 0.5 $

3. Calculate $\omega_{max}^2$:

   $ \omega_{max}^2 = \frac{2 \times 10 \text{ m/s}^2 \times 0.5}{0.1 \text{ m}} = \frac{10}{0.1} = 100 \text{ (rad/s)}^2 $

4. Find $\omega_{max}$:

   $ \omega_{max} = \sqrt{100} = 10 \text{ rad/s} $

5. Calculate Maximum Induced EMF ($E_{max}$):

   Substitute the values into the EMF formula:

   $ E_{max} = \frac{1}{2} \times (2 \text{ T}) \times (0.1 \text{ m})^2 \times (10 \text{ rad/s}) $

   $ E_{max} = 1 \times 0.01 \times 10 = 0.1 \text{ V} $

6. Convert Voltage to Millivolts:

   $ E_{max} = 0.1 \text{ V} \times \frac{1000 \text{ mV}}{1 \text{ V}} = 100 \text{ mV} $

The maximum induced EMF is 100 mV.