A solid metal cube of edge length 2 cm is moving in a positive y direction at a constant speed of 6 m/s. There is a uniform magnetic field of 0.1 T in the positive z-direction. The potential difference between the two faces of the cube perpendicular to the x-axis is:
A) 6 mV
B) 1 mV
C) 12 mV
D) 2 mV

Given Data:
• Edge length of the metal cube = 2 cm = 0.02 m
• Speed of the cube in the positive y-direction = 6 m/s
• Magnetic field B=0.1 TB = 0.1 \, \text{T} in the positive z-direction
• We need to find the potential difference between the two faces of the cube perpendicular to the x-axis.
Concept:
The potential difference across the metal cube is induced due to its motion through the magnetic field. This phenomenon is described by Faraday’s Law of Induction, and the formula for the induced EMF (electromotive force) is:
E=B⋅v⋅l\mathcal{E} = B \cdot v \cdot l
Where:
• E\mathcal{E} is the induced EMF (potential difference),
• BB is the magnetic field strength,
• vv is the velocity of the cube,
• ll is the length of the side of the cube (since the two faces of the cube perpendicular to the x-axis are the ones where the potential difference is measured).
Step 1: Apply the formula for induced EMF
The induced EMF is given by the equation:
E=B⋅v⋅l\mathcal{E} = B \cdot v \cdot l
Substitute the given values:
• B=0.1 TB = 0.1 \, \text{T}
• v=6 m/sv = 6 \, \text{m/s}
• l=0.02 ml = 0.02 \, \text{m}
E=0.1 T×6 m/s×0.02 m\mathcal{E} = 0.1 \, \text{T} \times 6 \, \text{m/s} \times 0.02 \, \text{m} E=0.012 V=12 mV\mathcal{E} = 0.012 \, \text{V} = 12 \, \text{mV}
Conclusion:
The potential difference between the two faces of the cube perpendicular to the x-axis is 12 mV.
Therefore, the correct answer is:
C. 12 mV.