A proton (or charged particle) moving with ve | vedclass

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(C) The total force acting on a charged particle moving in both electric and magnetic fields is given by the Lorentz force equation: $\overrightarrow{

Vedclass · Accepted Answer

$E$,$B$,and $v$ are mutually perpendicular and $v = \frac{E}{B}$

Vedclass · Answer

(C) The total force acting on a charged particle moving in both electric and magnetic fields is given by the Lorentz force equation: $\overrightarrow{F} = q(\overrightarrow{E} + \overrightarrow{v} 	imes \overrightarrow{B})$.For the proton to move undeflected,the net force must be zero,meaning $\overrightarrow{F} = 0$.This implies $q\overrightarrow{E} = -q(\overrightarrow{v} 	imes \overrightarrow{B})$,or $\overrightarrow{E} = -(\overrightarrow{v} 	imes \overrightarrow{B})$.This condition is satisfied when $\overrightarrow{E}$,$\overrightarrow{B}$,and $\overrightarrow{v}$ are mutually perpendicular to each other.In this configuration,the magnitude of the electric force $|F_e| = qE$ must equal the magnitude of the magnetic force $|F_m| = qvB$.Therefore,$qE = qvB$,which simplifies to $v = \frac{E}{B}$.

$A$ proton (or charged particle) moving with velocity $v$ is acted upon by an electric field $E$ and a magnetic field $B$. The proton will move undeflected if

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