Question

A conducting rod of 1m length rotating with a frequency of 50 rev/sec about its one of end inside the uniform magnetic field of 6.28mt. The value of induced emf

Answer 1

the value of induced emf is 25rev/sev

Answer 2

Answer:

The value of induced emf in the rod is 0.98 V

Explanation:

Given:

Length of the conducting rod, l=1 m

frequency of the rotation of the rod, $\nu$ =50 rev/s

Magnitude of the magnetic field, $B=6.28\times10^{-3}\ \rm T$

We know that the magnitude of the emf generated across the rod as it moves at right angles to the magnetic field is given by:

$de=Bvdr$

Where

• de is the emf induced in the small length dl
• b is the magnetic Field
• dr is them small length of the rod

Integrating the emf over the entire length we have

$e=\int Bvdl$

Also we know that $v=\omega r$

$e=\int B\omega rdr$

$e=\dfrac{B\omega r^2}{2}\\e=\dfrac{6.28\times 10^{-3}\times2\pi \times50\times1}{2}\\e=0.98\ \rm V$

Hence the emf induced in the is calculated.