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Step-by-Step Solution
1. Understanding the Problem
A charged particle with charge $q$ enters a region where there are uniform electric and magnetic fields, $ \vec{E} $ and $ \vec{B} $, respectively. These fields are perpendicular (orthogonal) to each other, and the velocity $ \vec{v} $ of the charged particle is perpendicular to both $ \vec{E} $ and $ \vec{B} $. The particle emerges from this region without any change in the magnitude or direction of $ \vec{v} $. We need to determine the relation between $ \vec{v} $, $ \vec{E} $, and $ \vec{B} $.
2. Condition for Unchanged Velocity
If the velocity $ \vec{v} $ of the particle remains unchanged as it moves through perpendicular electric and magnetic fields, the net force on the particle must be zero. The forces in play are:
Electric force: $ \vec{F}_\text{electric} = q \,\vec{E} $
Magnetic force: $ \vec{F}_\text{magnetic} = q\,(\vec{v} \times \vec{B}) $
For no change in $ \vec{v} $, these forces must cancel out:
$ q \,\vec{E} + q \,(\vec{v} \times \vec{B}) = 0 \quad \Rightarrow \quad \vec{E} + (\vec{v} \times \vec{B}) = 0. $
3. Relationship Between $ \vec{v} $, $ \vec{E} $, and $ \vec{B} $
Because $ \vec{v} $ is perpendicular to $ \vec{B} $ and $ \vec{E} $ is also perpendicular to $ \vec{B} $, we can deduce that:
$ \vec{v} \times \vec{B} $ must be in the opposite direction to $ \vec{E} $. This implies that the magnitudes satisfy:
$ q E = q\,v\,B \quad \Longrightarrow \quad v = \frac{E}{B}. $
4. Confirming the Correct Direction
The cross product $ \vec{E} \times \vec{B} $ is a vector perpendicular to both $ \vec{E} $ and $ \vec{B} $. Since $ \vec{v} $ itself is perpendicular to both $ \vec{E} $ and $ \vec{B} $, it must be parallel or antiparallel to $ \vec{E} \times \vec{B} $. Therefore, a logical form is:
$ \vec{v} = \alpha \,\bigl( \vec{E} \times \vec{B} \bigr), $
for some scalar $ \alpha $. We now determine $ \alpha $ by matching magnitudes.
5. Matching Magnitudes
The magnitude of $ \vec{E} \times \vec{B} $ is $ E\,B\,\sin 90^\circ = E\,B $, since $ \vec{E} $ and $ \vec{B} $ are perpendicular. Hence,
$ \bigl|\vec{E} \times \vec{B}\bigr| = E\,B. $
We want the resulting velocity vector $ \vec{v} $ to have magnitude $ v = \frac{E}{B} $. Therefore,
$ \bigl|\vec{v}\bigr| = \bigl|\alpha \,(\vec{E} \times \vec{B})\bigr| = \alpha \,E\,B. $
Since we require $ \bigl|\vec{v}\bigr| = \frac{E}{B} $, it follows that:
$ \alpha\,E\,B = \frac{E}{B} \quad \Longrightarrow \quad \alpha = \frac{1}{B^2}. $
6. Conclusion
Putting it all together, we get:
$ \vec{v} = \frac{\vec{E} \times \vec{B}}{B^2}. $
This is the velocity vector that remains perpendicular to both $ \vec{E} $ and $ \vec{B} $ and maintains its magnitude as the particle traverses the region with no net force acting on it.
