A charged particle moves with constant veloci | vedclass

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Question

(A, C, D) The net force on a charged particle moving with velocity $\vec{v}$ in the presence of electric field $\vec{E}$ and magnetic field $\vec{B}$

Vedclass · Accepted Answer

$\vec{E} = 0, \vec{B} 
eq 0$

Vedclass · Answer

(A, C, D) The net force on a charged particle moving with velocity $\vec{v}$ in the presence of electric field $\vec{E}$ and magnetic field $\vec{B}$ is given by the Lorentz force: $\vec{F} = q(\vec{E} + \vec{v} 	imes \vec{B})$.For the velocity to remain constant, the net force must be zero $(\vec{F} = 0)$.Case $(i)$: If $\vec{E} = 0$ and $\vec{B} = 0$, then $\vec{F} = 0$. The particle moves with constant velocity.Case (ii): If $\vec{E} 
eq 0$ and $\vec{B} 
eq 0$, the forces can cancel each other if $\vec{E} = -(\vec{v} 	imes \vec{B})$. This is possible.Case (iii): If $\vec{E} = 0$ and $\vec{B} 
eq 0$, the force is $\vec{F} = q(\vec{v} 	imes \vec{B})$. If $\vec{v}$ is parallel or anti-parallel to $\vec{B}$, then $\vec{v} 	imes \vec{B} = 0$, so $\vec{F} = 0$. The particle moves with constant velocity.Case (iv): If $\vec{E} 
eq 0$ and $\vec{B} = 0$, the force is $\vec{F} = q\vec{E}$. For $\vec{F} = 0$, $\vec{E}$ must be zero, which contradicts the assumption. Thus, this case is not possible.Therefore, cases $(A)$, $(C)$, and $(D)$ are possible.

$A$ charged particle moves with constant velocity in a region where no effect of gravity is felt, but an electrostatic field $\vec{E}$ together with a magnetic field $\vec{B}$ may be present. Which of the following cases are possible?

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