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NCERT Exemplar · Q10

Q.A charged particle would continue to move with a constant velocity in a region wherein,

(a) E=0, B≠0E = 0,\ B \neq 0.
(b) E≠0, B≠0E \neq 0,\ B \neq 0.
(c) E≠0, B=0E \neq 0,\ B = 0.
(d) E=0, B=0E = 0,\ B = 0.
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Constant velocity needs zero net Lorentz force, q(E⃗+v⃗×B⃗)=0q(\vec{E}+\vec{v}\times\vec{B})=0. That is possible for (a) E=0,B≠0E=0,B\neq0 with v⃗∥B⃗\vec{v}\parallel\vec{B}; for (b) E≠0,B≠0E\neq0,B\neq0 with E⃗=−v⃗×B⃗\vec{E}=-\vec{v}\times\vec{B}; and for (d) E=0,B=0E=0,B=0 trivially. It is impossible for (c) E≠0,B=0E\neq0,B=0.

Condition for constant velocity

Constant velocity means zero acceleration, hence zero net force. The only force acting is the Lorentz force, so

q(E⃗+v⃗×B⃗)=0⟹E⃗+v⃗×B⃗=0.q(\vec{E}+\vec{v}\times\vec{B})=0\quad\Longrightarrow\quad \vec{E}+\vec{v}\times\vec{B}=0.

(a) E=0, B≠0E=0,\ B\neq0. The condition reduces to v⃗×B⃗=0\vec{v}\times\vec{B}=0, satisfied when v⃗\vec{v} is parallel (or anti-parallel) to B⃗\vec{B}. The magnetic force is then zero and the velocity stays unchanged. Possible.

(b) E≠0, B≠0E\neq0,\ B\neq0. Possible when the electric and magnetic forces exactly cancel: E⃗=−v⃗×B⃗\vec{E}=-\vec{v}\times\vec{B} (with E⃗⊥B⃗\vec{E}\perp\vec{B} and speed v=E/Bv=E/B). This is precisely the working principle of a velocity selector. Possible. …

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