A proton moving with a constant velocity pass | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The net force on a charged particle moving with constant velocity must be zero. The Lorentz force is given by $\vec{F} = q(\vec{E} + \vec{v} 	ime

Vedclass · Accepted Answer

$(A), (B)$ and $(D)$ only

Vedclass · Answer

(C) The net force on a charged particle moving with constant velocity must be zero. The Lorentz force is given by $\vec{F} = q(\vec{E} + \vec{v} 	imes \vec{B})$. For $\vec{F} = 0$,we have $\vec{E} + \vec{v} 	imes \vec{B} = 0$.Case $(A)$: If $\vec{E} = 0$ and $\vec{B} = 0$,then $\vec{F} = 0$. The velocity remains constant.Case $(B)$: If $\vec{E} = 0$ and $\vec{B} 
eq 0$,the particle can move with constant velocity if $\vec{v}$ is parallel or anti-parallel to $\vec{B}$ (since $\vec{v} 	imes \vec{B} = 0$).Case $(C)$: If $\vec{E} 
eq 0$ and $\vec{B} = 0$,the electric force $q\vec{E}$ would cause acceleration,so the velocity cannot remain constant. Thus,$(C)$ is not possible.Case $(D)$: If $\vec{E} 
eq 0$ and $\vec{B} 
eq 0$,the particle can move with constant velocity if $\vec{E} = -(\vec{v} 	imes \vec{B})$. This is possible (e.g.,in a velocity selector).Therefore,cases $(A), (B),$ and $(D)$ are possible.

Similar Questions

If the temperature of the cold junction of a thermocouple is lowered,then the neutral temperature

The dimensional formula of $\frac{1}{2} \mu_0 H^2$ (where $\mu_0$ is the permeability of free space and $H$ is the magnetic field intensity) is:

$A$ thin,straight conductor lies along the axis of a hollow conductor of radius $R$. The two carry equal currents in the same direction. The magnetic field $B$ is plotted against the distance $r$ from the axis. Which of the following best represents the resulting curve?

Vedclass Test Series

Exam Paper Generator

Online Exam Module