Question

A metal rod of length $2m$ is rotating with an angular velocity of $100$ $rad{s^{ - 1}}$ in a plane perpendicular to a uniform magnetic field of $0.3T$. The potential difference between the ends of the rod is:
A) 30V
B) 40V
C) 60V
D) 600V

Answer 1

Hint: The voltage produced due to change in the magnetic flux linked with the metal rod is given by Faraday's law of electromagnetic induction which states that the voltage induced is directly proportional to rate of change of magnetic flux. The expression is given by –
The emf induced,
$E = - N\dfrac{{d\phi }}{{dt}}$
where, N= number of turns of the coil, $\phi $= Magnetic flux.

Complete step by step solution:
Electromotive force (emf) is energy per unit electric charge that is released by a source of energy. Emf can also be written as,
$E = - N\dfrac{{d\phi }}{{dt}}$
where, N= number of turns, $\phi $= Magnetic flux.
Again, $\phi = BA\cos \theta $,
Where, B= External magnetic field, $\theta $=Angle between the coil and the magnetic field, and A=Area of the coil.
Thus, we get,
$E = - N\dfrac{d}{{dt}}\left( {BA} \right)$
But, we have, N= 1 and B is constant. So, from the above equation we get,
$E = - B\dfrac{{dA}}{{dt}}$
$ \Rightarrow E = Blv$
Where, B= Magnetic field, l= length of the coil, and v= velocity of the movement of the coil over the magnetic field.
Given that the initial emf is zero and the final emf is equal to $E = Blv$ , their average is given by –
Average EMF, ${E_{avg}} = \dfrac{{0 + E}}{2} = \dfrac{1}{2}Blv$
Now, if we consider the angular velocity $\omega $, then we have, $v = l\omega $, thus the equation can be rewritten as,
${E_{avg}} = \dfrac{1}{2}B{l^2}\omega $
Given, in the question, B= 0.3 T, l= 2m, and $\omega $=100 $rad{s^{ - 1}}$.
Substituting, we get –
${E_{avg}} = \dfrac{1}{2} \times 0 \cdot 3 \times {2^2} \times 100$
$ \Rightarrow {E_{avg}} = \dfrac{1}{2} \times 0 \cdot 3 \times 4 \times 100$
$ \Rightarrow {E_{avg}} = \dfrac{1}{2} \times 120$
$\therefore {E_{avg}} = 60V$

Thus, the potential difference between the end of the rod is 60V.

Note: Here, we have considered the emf as positive value since we are only interested in the magnitude in this problem. However, in actuality, the direction of the emf is opposite to that of the magnetic field and hence, the sign is negative. The actual voltage induced is equal to – 60V.