Question

A metallic rod of 2 m length is rotated with a frequency 100 Hz about an axis passing through the centre of the circular ring of radius 2 m. A constant magnetic field 2 T is applied parallel to the axis and perpendicular to the length of the rod. The emf developed across the ends of the rod is:

• A. 800 \(\pi\) volt
• B. 1600 \(\pi\) volt
• C. 1600 volt
• D. 400 \(\pi\) volt

Hint:

For calculating the emf in a rotating conductor within a magnetic field, use the formula \( \mathcal{E} = B \omega l^2 \), where \( \omega = 2\pi f \) relates the frequency to the angular velocity.

Answer 1

The Correct Option is A

The emf induced in a rotating rod placed in a magnetic field is given by the formula: \[ \mathcal{E} = B \omega l^2 \] Where: - \( \mathcal{E} \) is the induced emf, - \( B \) is the magnetic field strength, - \( \omega \) is the angular velocity, - \( l \) is the length of the rod. The angular velocity \( \omega \) is related to the frequency \( f \) by: \[ \omega = 2\pi f \] Now, given: - \( B = 2 \, \text{T} \), - \( f = 100 \, \text{Hz} \), - \( l = 2 \, \text{m} \), First, calculate the angular velocity: \[ \omega = 2 \pi \times 100 = 200\pi \, \text{rad/s} \] Now substitute the values into the formula for emf: \[ \mathcal{E} = 2 \times 200\pi \times 2^2 = 2 \times 200\pi \times 4 = 1600 \pi \, \text{volt} \] Thus, the emf developed across the ends of the rod is \( 1600 \pi \) volts.