# Magnetic Field and Induced Current in a rod

So, I know that a changing magnetic flux will create an e.m.f. And since magnetic flux is dependent on the magnetic field strength and area, a changing area will induce current. However, if I have a rotating rod in a magnetic field, why is current induced? I mean, it's not like the rod changes in area or volume. And since the magnetic field is uniform and constant, the magnetic flux shouldn't be changing as I rotate the rod in the magnetic field right? Let's consider what happens to charged particles in the conducting rod. I'll use a cylindrical (r, z, theta) coordinate system, with the origin located at the rod's fixed point, and the magnetic field pointing in the z-direction.

The charges in the conductor experience a $$q \vec{v} \times \vec{B}$$ force moving positive charges in one direction towards one end of the rod, and negative charges to the other. This charge separation sets up an electric field, and an equilibrium is reached where the net force on a given charged particle:

$$\vec{F} = q \left(\vec{E} + \vec{v} \times \vec{B}\right) = 0$$

Or,

$$E = - \omega r B$$ (where r radial position along the rod)

This results in a potential difference along the rod of:

$$V = \int \vec{E} \cdot d\vec{l} = -\frac{B\omega L^2}{2}$$

As this potential difference is a result of the motion of the conductor, it is referred to as a motional emf.

Now, perhaps the source of your confusion is in understanding the application of Faraday's law. You're quite right in saying the emf in any closed circuit can be related to the change of magnetic flux through that circuit according to:

$$\epsilon = - \frac{d \phi}{dt}$$.

I added my emphasis as in your original question there is no close circuit (loop). If we create a loop that is formed by the a line along the rod in its current position, the rod's original position, and the chord that the far end of the rod follows, then the area of that loop as a function of time is given by:

$$A(t) = \frac{1}{2}L^2 \theta(t)$$,

Therefore

$$\epsilon = - \frac{d \phi}{dt} = -\frac{1}{2}L^2\omega B$$.

Alternatively, one can calculate the motional emf as the integral around that loop of the force per unit charge (work done per unit charge):

$$\epsilon = \oint \left(\vec{v} \times \vec{B}\right) \cdot d \vec{l}$$,

which due to the fact the only section of the loop with non-zero velocity is the rod itself gives the same answer.

There is a concept called motional emf which says motion in magnetic field (in specific configuration) induces a voltage. The voltage induced across the ends of element of length dl is given by $$dV=(\vec v × \vec B).\vec{ dl}$$. Here $$\vec v$$ the velocity . So you can integrate the result for a rotating rod.