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: For a charged particle to move without acceleration, \[ \vec F=q(\vec E+\vec v\times\vec B)=0. \] Hence, \[ \vec E+\vec v\times\vec B=0. \] Therefore, \[ \vec v\times\vec B=-\vec E. \]

Step 1: Determine the direction of velocity. Given, \[ \vec E=2\times10^{3}\hat i, \qquad \vec B=5\times10^{-2}\hat j. \] We need \[ \vec v\times\hat j=-\hat i. \] Using vector products, \[ \hat k\times\hat j=-\hat i. \] Therefore, \[ \vec v \parallel \hat k. \]

Step 2: Find the magnitude of velocity. Since \(\vec v\) is perpendicular to \(\vec B\), \[ E=vB. \] Hence, \[ v=\frac{E}{B}. \] Substituting the given values, \[ v = \frac{2\times10^{3}} {5\times10^{-2}}. \] \[ v = \frac{2}{5}\times10^{5}. \] \[ v = 0.4\times10^{5}. \] \[ v = 4\times10^{4}\ \text{m s}^{-1}. \]

Step 3: Write the velocity vector. \[ \boxed{\vec v=4\times10^{4}\,\hat k\ \text{m s}^{-1}} \] \[ \boxed{\text{Answer = (C)}} \]