Question

A simple pendulum made of mass 10 g and a metallic wire of length 10 cm is suspended vertically in a uniform magnetic field of 2 T. The magnetic field direction is perpendicular to the plane of oscillations of the pendulum. If the pendulum is released from an angle of 60° with vertical, then maximum induced EMF between the point of suspension and point of oscillation is_______ mV. (Take g = 10 m/s²)}

Hint:

The induced EMF is zero at the extreme positions (where $\omega = 0$) and maximum at the equilibrium position (where $\omega$ is maximum).

Answer 1

Correct Answer: 100

Step 1: Understanding the Concept:
A metallic wire oscillating in a magnetic field acts as a conductor rotating about a fixed point. The induced EMF between the pivot and the tip is given by the formula for a rotating rod, but with the instantaneous angular velocity of the pendulum. The maximum EMF occurs when the velocity (and hence angular velocity) is maximum, which is at the mean position.

Step 2: Key Formula or Approach:
1. Induced EMF in a rotating rod: \(\varepsilon = \frac{1}{2}BL^2\omega\).
2. Conservation of energy to find maximum angular velocity: \(mgL(1 - \cos\theta) = \frac{1}{2}mv^2 = \frac{1}{2}m(L\omega_{max})^2\).
Step 3: Detailed Explanation:
First, find \(\omega_{max}\) at the lowest point:
\[ gL(1 - \cos 60^\circ) = \frac{1}{2}L^2\omega_{max}^2 \] \[ 10 \times 0.1 \times (1 - 0.5) = \frac{1}{2} \times (0.1)^2 \times \omega_{max}^2 \] \[ 0.5 = 0.005 \times \omega_{max}^2 \] \[ \omega_{max}^2 = 100 \implies \omega_{max} = 10 \text{ rad/s} \] Now, calculate the maximum induced EMF: \[ \varepsilon_{max} = \frac{1}{2} B L^2 \omega_{max} \] \[ \varepsilon_{max} = \frac{1}{2} \times 2 \times (0.1)^2 \times 10 \] \[ \varepsilon_{max} = 1 \times 0.01 \times 10 = 0.1 \text{ V} \] To convert to mV: \[ 0.1 \text{ V} \times 1000 = 100 \text{ mV} \]
Step 4: Final Answer:
The maximum induced EMF is 100 mV.