Question

A charged particle moves in a region where there is a uniform electric field \[ \vec E = 2\times10^{3}\,\hat i \;\text{N C}^{-1} \] and a uniform magnetic field \[ \vec B = 5\times10^{-2}\,\hat j \;\text{T}. \] The velocity of the particle (in m s\(^{-1}\)) if the particle moves without any acceleration is

• A. \(5\times10^{4}\,\hat k\)
• B. \(2\times10^{3}\,\hat k\)
• C. \(4\times10^{4}\,\hat k\)
• D. \(4\times10^{4}(-\hat k)\) \bigskip

Hint:

For a charged particle moving undeflected in crossed electric and magnetic fields, \[ qE=qvB. \] Therefore, \[ v=\frac{E}{B}. \] The direction is obtained from \[ \vec v\times\vec B=-\vec E. \]

Answer 1

The Correct Option is C

Concept: No acceleration: \(q(\vec{E} + \vec{v}\times\vec{B}) = 0 \Rightarrow \vec{v}\times\vec{B} = -\vec{E}\). Solve for \(\vec{v}\) direction and magnitude.

Step 1: \(\vec{E} = 2\times10^3\hat{i}\), \(\vec{B} = 5\times10^{-2}\hat{j}\). \(\vec{v}\times\hat{j} = -\hat{i} \Rightarrow \vec{v} \parallel \hat{k}\). \(E = vB \Rightarrow v = 2\times10^3 / 5\times10^{-2} = 4\times10^4\) m/s. \(\vec{v} = 4\times10^4\hat{k}\) m/s.

Step 2: Write the final answer. \(\boxed{\vec v=4\times10^{4}\,\hat k\ \text{m s}^{-1}}\)