Question

The rod PQ slides along 2 parallel rails as shown in the figure. It has a length of 20 cm and is perpendicular to the 2 rails. It performs simple harmonic motion with amplitude 5 cm and frequency 10 Hz. The magnetic field is 10^-4T and is directed perpendicular to the plane of paper. What is the peak induced electro-magnetic force?

• A. 2\(\pi\)x10^-7 V
• B. 4π²x10^-3 V
• C. 2\(\pi\)x10^-5 V
• D. 4\(\pi\)x10^-5 V
• E. \(\pi\)²x10^-4 V

Answer 1

The Correct Option is C

Given:

• Length of rod (PQ), \( L = 20 \, \text{cm} = 0.2 \, \text{m} \)
• Amplitude of SHM, \( A = 5 \, \text{cm} = 0.05 \, \text{m} \)
• Frequency of SHM, \( f = 10 \, \text{Hz} \)
• Magnetic field, \( B = 10^{-4} \, \text{T} \) (perpendicular to plane)

Step 1: Determine Maximum Velocity (\( v_{\text{max}} \))

For simple harmonic motion, the maximum velocity is given by:

\[ v_{\text{max}} = A \omega \]

where angular frequency \( \omega = 2\pi f \).

Substitute values:

\[ v_{\text{max}} = 0.05 \times (2\pi \times 10) = \pi \, \text{m/s} \]

Step 2: Calculate Peak Induced EMF (\( \mathcal{E}_{\text{max}} \))

The induced EMF in a moving rod is:

\[ \mathcal{E} = B L v \]

At peak velocity, the peak EMF is:

\[ \mathcal{E}_{\text{max}} = B L v_{\text{max}} \]

Substitute known values:

\[ \mathcal{E}_{\text{max}} = 10^{-4} \times 0.2 \times \pi = 2\pi \times 10^{-5} \, \text{V} \]

Conclusion:

The peak induced electro-magnetic force is \( 2\pi \times 10^{-5} \, \text{V} \).

Answer: \(\boxed{C}\)

Answer 2

The rod has a length of \( 20 \, \text{cm} \), an amplitude of \( 5 \, \text{cm} \), and a frequency of \( 10 \, \text{Hz} \). The magnetic field is \( B = 10^{-4} \, \text{T} \) and is directed perpendicular to the plane of the paper.

Step 1: Recall Faraday's Law of Electromagnetic Induction 

The induced EMF (\( \mathcal{E} \)) in a conductor moving through a magnetic field is given by: \[ \mathcal{E} = B \cdot l \cdot v, \] where: - \( B \) is the magnetic field strength, - \( l \) is the length of the conductor, - \( v \) is the velocity of the conductor. For SHM, the velocity of the rod as a function of time is: \[ v(t) = A \omega \cos(\omega t), \] where: - \( A \) is the amplitude of the motion, - \( \omega \) is the angular frequency (\( \omega = 2\pi f \)), - \( f \) is the frequency of the motion. The peak velocity (\( v_{\text{peak}} \)) occurs when \( \cos(\omega t) = 1 \): \[ v_{\text{peak}} = A \omega. \] Step 2: Substitute Known Values 

1. Length of the rod (\( l \))**: \[ l = 20 \, \text{cm} = 0.2 \, \text{m}. \] 2. Amplitude (\( A \)): \[ A = 5 \, \text{cm} = 0.05 \, \text{m}. \] 3. Frequency (\( f \)): \[ f = 10 \, \text{Hz}. \] 4. Angular frequency (\( \omega \)): \[ \omega = 2\pi f = 2\pi \times 10 = 20\pi \, \text{rad/s}. \] 5. Peak velocity (\( v_{\text{peak}} \)): \[ v_{\text{peak}} = A \omega = 0.05 \times 20\pi = \pi \, \text{m/s}. \] 6. Magnetic field (\( B \)): \[ B = 10^{-4} \, \text{T}. \] Step 3: Calculate the Peak Induced EMF 

Using the formula for induced EMF: \[ \mathcal{E}_{\text{peak}} = B \cdot l \cdot v_{\text{peak}}. \] Substitute the values: \[ \mathcal{E}_{\text{peak}} = 10^{-4} \cdot 0.2 \cdot \pi = 2 \times 10^{-5} \pi \, \text{V}. \] Simplify: \[ \mathcal{E}_{\text{peak}} = 2\pi \times 10^{-5} \, \text{V}. \] Final Answer: The peak induced electromotive force is: \[ \boxed{2\pi \times 10^{-5} \, \text{V}}, \] which corresponds to option \( \mathbf{(C)} \).