Given data

• Side of cube: $a = 15\text{ cm} = 0.15\text{ m}$
• Velocity: $\vec v = 2\,\text{m s}^{-1}\,\hat j$
• Magnetic field: $\vec B = 0.5\,\text{T}\,\hat k$

Direction of magnetic force on charges

A charge inside the moving conductor experiences magnetic force $\vec F = q(\vec v \times \vec B).$

Here, $\vec v \times \vec B = (2\hat j) \times (0.5\hat k) = 1\hat i.$

So positive charges are pushed along $+x$ direction.

Hence:

• face at larger $x$ becomes higher potential
• opposite face becomes lower potential

Electric field developed in equilibrium

In equilibrium, electric force balances magnetic force: $qE = qvB$ $E = vB$

Therefore, $E = 2 \times 0.5 = 1\,\text{V/m}.$

Potential difference between opposite faces

Separation between the two faces is the side of cube: $d = 0.15\,\text{m}.$

So, $V = Ed = 1 \times 0.15 = 0.15\,\text{V}.$

Convert to mV

$0.15\,\text{V} = 150\,\text{mV}.$

Final answer

The potential difference between the higher and lower potential faces is $\boxed{150\text{ mV}}.$