A metallic cube of side 15 cm moving along $$y$$-axis at a uniform velocity of 2 m s$$^{-1}$$. In a region of uniform magnetic field of magnitude 0.5 T directed along $$z$$-axis. In equilibrium the potential difference between the faces of higher and lower potential developed because of the motion through the field will be _______ mV.

A metallic cube of side 15 cm moves along the y-axis at 2 m/s in a magnetic field of 0.5 T along the z-axis. We need to find the potential difference between the faces.

To begin,

When a conductor moves through a magnetic field, the free charges inside experience a Lorentz force. The velocity is along $$\hat{j}$$ (y-axis) and the magnetic field is along $$\hat{k}$$ (z-axis). The magnetic force on a positive charge is:

$$ \vec{F} = q\vec{v} \times \vec{B} = q(v\hat{j}) \times (B\hat{k}) $$

Using $$\hat{j} \times \hat{k} = \hat{i}$$:

$$ \vec{F} = qvB\hat{i} $$

So the force on positive charges is in the $$+x$$ direction. This causes charge separation along the x-axis.

Next,

The charges separate until the electric field from the separation balances the magnetic force. The induced EMF across the length $$l$$ in the x-direction is:

$$ \mathcal{E} = Bvl $$

where $$l = 15$$ cm = 0.15 m is the side of the cube (the dimension along which the potential difference develops, i.e., the x-direction).

$$ \mathcal{E} = 0.5 \times 2 \times 0.15 = 0.15\;\text{V} = 150\;\text{mV} $$

The potential difference between the faces is 150 mV.