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.
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.
Correct Answer: 150
Solution and Explanation
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 \( 150 \, \text{mV} \).
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