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(B) The net Lorentz force on the particle is $\vec{F} = q(\vec{E} + \vec{v} 	imes \vec{B})$.Since the particle passes through the region without any

Vedclass · Accepted Answer

$\vec{v} = \frac{(\vec{E} 	imes \vec{B})}{B^2}$

Vedclass · Answer

(B) The net Lorentz force on the particle is $\vec{F} = q(\vec{E} + \vec{v} 	imes \vec{B})$.Since the particle passes through the region without any change in velocity,the net force must be zero,so $\vec{E} + \vec{v} 	imes \vec{B} = 0$,which implies $\vec{E} = -(\vec{v} 	imes \vec{B}) = \vec{B} 	imes \vec{v}$.Taking the cross product with $\vec{B}$ on both sides: $\vec{E} 	imes \vec{B} = (\vec{B} 	imes \vec{v}) 	imes \vec{B}$.Using the vector triple product identity $(\vec{A} 	imes \vec{B}) 	imes \vec{C} = \vec{B}(\vec{A} \cdot \vec{C}) - \vec{A}(\vec{B} \cdot \vec{C})$,we get $\vec{E} 	imes \vec{B} = \vec{v}(\vec{B} \cdot \vec{B}) - \vec{B}(\vec{v} \cdot \vec{B})$.Since $\vec{v} \perp \vec{B}$,$\vec{v} \cdot \vec{B} = 0$,so $\vec{E} 	imes \vec{B} = \vec{v} B^2$.Therefore,$\vec{v} = \frac{\vec{E} 	imes \vec{B}}{B^2}$.

$A$ charged particle with charge $q$ enters a region of constant,uniform,and mutually orthogonal fields $\vec{E}$ and $\vec{B}$ with a velocity $\vec{v}$ perpendicular to both $\vec{E}$ and $\vec{B}$,and comes out without any change in the magnitude or direction of $\vec{v}$. Then:

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