Question

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

• A. $\overset{\rightarrow}{v} = \overset{\rightarrow}{E}\mspace{6mu} \times \overset{\rightarrow}{B}/B^{2}$
• B. $\overset{\rightarrow}{v} = \overset{\rightarrow}{B}\mspace{6mu} \times \overset{\rightarrow}{E}/B^{2}$
• C. $\overset{\rightarrow}{v} = \overset{\rightarrow}{E}\mspace{6mu} \times \overset{\rightarrow}{B}/E^{2}$
• D. $\overset{\rightarrow}{v} = \overset{\rightarrow}{B}\mspace{6mu} \times \overset{\rightarrow}{E}/E^{2}$

Answer 1

Answer: (A)

$\overset{\rightarrow}{v} = \overset{\rightarrow}{E}\mspace{6mu} \times \overset{\rightarrow}{B}/B^{2}$

Answer 2

$\overset{\rightarrow}{F_{net}}$= $\overset{\rightarrow}{F_{e}}$ + $\overset{\rightarrow}{F_{m}}$ 0 = q$\overset{\rightarrow}{E}$ + q($\overset{\rightarrow}{v} \times \overset{\rightarrow}{B}$) Ž $\overset{\rightarrow}{E}$ = $\overset{\rightarrow}{B}\mspace{6mu} \times \overset{\rightarrow}{v}$

Ž $\overset{\rightarrow}{E}\mspace{6mu} \times \overset{\rightarrow}{B}$ = $\left( \overset{\rightarrow}{B}\mspace{6mu} \times \overset{\rightarrow}{v} \right)$ × $\overset{\rightarrow}{B}$ = $\overset{\rightarrow}{v}$ – $\overset{\rightarrow}{B}$

Q B, E & v are mutually ^ as given Ž$\overset{\rightarrow}{v}$= $\frac{\overset{\rightarrow}{E}\mspace{6mu} \times \overset{\rightarrow}{B}}{B^{2}}$