If E and B are the magnitudes of electric and | vedclass

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Question

(A) The force on a charged particle moving with velocity $\vec{v}$ in an electric field $\vec{E}$ and magnetic field $\vec{B}$ is given by the Lorentz

Vedclass · Accepted Answer

$E=v B$

Vedclass · Answer

(A) The force on a charged particle moving with velocity $\vec{v}$ in an electric field $\vec{E}$ and magnetic field $\vec{B}$ is given by the Lorentz force equation: $\vec{F} = q(\vec{E} + \vec{v} 	imes \vec{B})$.For the particle to move with a uniform velocity, the net force must be zero, i.e.,$\vec{F} = 0$.Case $1$: If $E=0$ and $B=0$, the particle moves with uniform velocity (Newton's first law).Case $2$: If $E=0$ and $B 
eq 0$, the particle moves with uniform velocity if it moves parallel or anti-parallel to the magnetic field $(\vec{v} 	imes \vec{B} = 0)$.Case $3$: If $E 
eq 0$ and $B 
eq 0$, the particle moves with uniform velocity if the electric force and magnetic force cancel each other, which occurs when $\vec{E} = -(\vec{v} 	imes \vec{B})$. In terms of magnitudes, this implies $E = vB$ (where $\vec{v} \perp \vec{B}$ and $\vec{E} \perp \vec{v}$).Thus, the condition $E=vB$ is a valid possibility for uniform velocity.

If $E$ and $B$ are the magnitudes of electric and magnetic fields respectively in some region of space, then the possibilities for which a charged particle may move in that space with a uniform velocity of magnitude $v$ are

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