Question

A metallic cube of side 15 cm moving along y-axis at a uniform velocity of 2 ms^–1. In a region of uniform magnetic field of magnitude 0.5T 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.

Answer 1

Correct Answer: 150

Given: 

• Side of the cube (\( l \)) = \( 15 \, \text{cm} = 0.15 \, \text{m} \)
• Velocity (\( \vec{v} \)) = \( 2\hat{j} \, \text{m/s} \)
• Magnetic field (\( \vec{B} \)) = \( 0.5\hat{k} \, \text{T} \)

Step 1: Formula for Induced EMF

The induced electromotive force (EMF) or potential difference (\( V \)) across the faces of the cube due to its motion in the magnetic field is given by:

\[ V = B l v, \]

where:

• \( B \): Magnetic field strength
• \( l \): Length of the side of the cube perpendicular to both velocity and magnetic field
• \( v \): Velocity of the cube

Step 2: Substitute the Given Values

Substitute \( B = 0.5 \, \text{T} \), \( l = 0.15 \, \text{m} \), and \( v = 2 \, \text{m/s} \) into the formula:

\[ V = (0.5)(0.15)(2). \]

Simplify:

\[ V = 0.15 \, \text{V}. \]

Convert to millivolts:

\[ V = 150 \, \text{mV}. \]

Final Answer:

The potential difference developed between the faces is \( 150 \, \text{mV} \).