An electron is released from the origin at a place where a uniform electric field E and uniform magnetic field B exit along the negative y-axis and the negative z-axis respectively-

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Standard XII

Physics

Unit Vectors

An electron i...

Question

An electron is released from the origin at a place where a uniform electric field $\to E$ and uniform magnetic field $\to B$ exit along the negative y-axis and the negative z-axis respectively.

At time

t

the y-component of velocity of the electron becomes

u_{y} = \frac{E}{B} sin ω t

where

ω = \frac{e B}{m}

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t = π m / e B

the electron will have only x-component of velocity.

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t = \frac{2 π m}{3 e B}

, the y-component of velocity becomes zero.

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The displacement along y-axis is

\frac{2 E m}{e B^{2}}

when the velocity of electron becomes perpendicular to the y-axis

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Solution

The correct options are
A At time $t$ the y-component of velocity of the electron becomes $u_{y} = \frac{E}{B} sin ω t$ where $ω = \frac{e B}{m}$ .
B At $t = π m / e B$ the electron will have only x-component of velocity.

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An electron is released from the origin at a place where a uniform electric field →E and uniform magnetic field →B exit along the negative y-axis and the negative z-axis respectively.

The correct options are A At time t the y-component of velocity of the electron becomes uy=EBsinωt where ω=eBm. B At t=πm/eB the electron will have only x-component of velocity.

An electron is released from the origin at a place where a uniform electric field $\to E$ and uniform magnetic field $\to B$ exit along the negative y-axis and the negative z-axis respectively.

The correct options are
A At time $t$ the y-component of velocity of the electron becomes $u_{y} = \frac{E}{B} sin ω t$ where $ω = \frac{e B}{m}$ .
B At $t = π m / e B$ the electron will have only x-component of velocity.