A charged particle with charge q enters a region of constant, uniform and mutually orthogonal fields E and B with a velocity v perpendicular to both E and B, and comes out without any change in magnitude or direction of v. Then (a) v = B × E/E² (b) v = E × B/B² (c) v = B × E/B² (d) v = E × B/E²

Answer: (b). No deflection: E + v× B = 0 ⇒ E = B× v Cross with B: E× B = ( B× v)× B = B² v - ( v· B) B = B² v (since v⊥ B) v = ( E× B)/(B²) So the answer is (b).

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Moving Charge in a Magnetic Field

A charged particle with charge $q$ enters a region of constant, uniform and mutually orthogonal fields $\vec{E}$ and $\vec{B}$ with a velocity $\vec{v}$ perpendicular to both $\vec{E}$ and $\vec{B},$ and comes out without any change in magnitude or direction of $\vec{v}$ . Then

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Answer: (b)

No deflection: $\vec E + \vec v\times\vec B = 0 \Rightarrow \vec E = \vec B\times\vec v$
Cross with $\vec B$ : $\vec E\times\vec B = (\vec B\times\vec v)\times\vec B = B^2\vec v - (\vec v\cdot\vec B)\vec B = B^2\vec v$ (since $\vec v\perp\vec B$ )
$\vec v = \dfrac{\vec E\times\vec B}{B^2}$
So the answer is (b).

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