Question

A metal rod of length 2 m is rotated with a frequency of 60 rev/s about an axis passing through its centre and perpendicular to its length. A uniform magnetic field of 2T perpendicular to its plane of rotation is switched-on in the region. Calculate the e.m.f. induced between the centre and the end of the rod.

Hint:

The induced e.m.f. in a rotating rod in a magnetic field depends on the magnetic field strength, the frequency of rotation, and the length of the rod.

Answer 1

Given:
Magnetic field \( B = \SI{2}{\tesla} \)
Length of the rod \( L = \SI{2}{\metre} \)
Frequency of rotation \( f = \SI{60}{\rev\per\second} \) The angular velocity \( \omega \) is given by: \[ \omega = 2 \pi f \] Substituting the value of \( f = 60 \, \text{rev/s} \): \[ \omega = 2 \pi \times 60 = 120 \pi \, \text{rad/s} \] The induced e.m.f. \( \mathcal{E} \) is given by: \[ \mathcal{E} = \frac{1}{2} B L^2 \omega \] Substituting the known values: \[ \mathcal{E} = \frac{1}{2} \times 2 \times (2)^2 \times (2 \pi \times 60) \] \[ \mathcal{E} = \frac{1}{2} \times 2 \times 4 \times 120 \pi \] \[ \mathcal{E} = 480 \pi \, \text{V} \] Finally, calculating the value of \( \mathcal{E} \) using \( \pi \approx 3.1416 \): \[ \mathcal{E} = 480 \times 3.1416 \, \text{V} \approx 1.51 \times 10^3 \, \text{V} \] \boxed{\mathcal{E} = 1.51 \times 10^3 \, \text{V}}