A charged particle is released from rest in a region of steady and uniform electric and magnetic fields which are parallel to each other. The particle will move in a

straight line

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Solution

The correct option is C straight line
$\to E ∥ \to B$
$F_{E} = q \to E$
$F_B= q(\vec{v} \times \vec{B}$
given $\to v$ is parallel to $\to B$
$\Rightarrow F_{B} = 0$ since $sin θ = sin 0^{\circ} = 0$
$∴$ the charged particle will accelerate and move in a straight line.

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A charged particle is released from rest in a region of steady and uniform electric and magnetic fields which are parallel to each other. The particle will move in a

The correct option is C straight line→E∥→BFE=q→E$F_B= q(\vec{v} \times \vec{B}$given →v is parallel to →B⇒FB=0 since sinθ=sin0∘=0 ∴ the charged particle will accelerate and move in a straight line.

The correct option is C straight line
$\to E ∥ \to B$
$F_{E} = q \to E$
$F_B= q(\vec{v} \times \vec{B}$
given $\to v$ is parallel to $\to B$
$\Rightarrow F_{B} = 0$ since $sin θ = sin 0^{\circ} = 0$
$∴$ the charged particle will accelerate and move in a straight line.