Question

A charged particle travels along a straight line with a speed $\nu$ in a region where both electric field $\mathbf{E}$ and magnetic fields $\mathbf{B}$ are present. It follows that}

• A. $\mathbf{E} = \nu \mid \mathbf{B}$ and the two fields are parallel
• B. $\mathbf{E} = \nu \mid \mathbf{B}$ and the two fields are perpendicular
• C. $\mathbf{B} = \nu \mid \mathbf{E}$ and the two fields are parallel
• D. $\mathbf{B} = \nu \mid \mathbf{E}$ and the two fields are perpendicular

Hint:

In a velocity selector, $\mathbf{E}$, $\mathbf{B}$, and $\mathbf{v}$ are mutually perpendicular with $v = E/B$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For a charged particle to move undeflected in a straight line, the net force on it must be zero.
Step 2: Detailed Explanation:
The Lorentz force is $\mathbf{F} = q(\mathbf{E} + \mathbf{v} \times \mathbf{B})$. For no deflection, $\mathbf{F} = 0$, so $\mathbf{E} = -(\mathbf{v} \times \mathbf{B})$. This implies $E = vB$ and $\mathbf{E}$ is perpendicular to both $\mathbf{v}$ and $\mathbf{B}$. Since $\mathbf{v}$ is along the line of motion, $\mathbf{E}$ and $\mathbf{B}$ are perpendicular to each other.
Step 3: Final Answer:
Thus $\mathbf{E} = \nu \mid \mathbf{B}$ and the two fields are perpendicular.