A charged particle goes undeflected in a region of space containing an electric field E and a magnetic field of intensity B- Which of the following is possible-A- E- is parallel to B- but v- is perpendicular to E-B- E is parallel to B and v parallel to EC- E- is perpendicular to B and v is parallel to E - BD- E is perpendicualar to B and v is parallel to E - B

From Lorentz force, F_L = F_e + F_m F_e = electric force F_m = magnetic force F_L = q E + q (v×B) If nbsp;E|| B and v|| E, then v must be parallel to B. ∴F_L ...

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Byju's Answer

Standard XII

Physics

Desciption of Force

A charged par...

Question

A charged particle goes undeflected in a region of space containing an electric field $\to E$ and a magnetic field of intensity $\to B$ . Which of the following is possible?

\to E

is parallel to

\to B

but

\to v

is perpendicular to

\to E

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\to E

is parallel to

\to B

and

\to v

parallel to

\to E

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\to E

is perpendicular to

\to B

and

\to v

is parallel to

\to E \times \to B

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\to E

is perpendicualar to

\to B

and

\to v

is parallel to

\to E . \to B

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Solution

The correct options are
B $\to E$ is parallel to $\to B$ and $\to v$ parallel to $\to E$
C $\to E$ is perpendicular to $\to B$ and $\to v$ is parallel to $\to E \times \to B$
From Lorentz force,
$_{L} =_{e} +_{m}$
$_{e}$ = electric force
$_{m}$ = magnetic force
$_{L} = q \to E + q (\to v \times \to B)$

If $\to E | | \to B$ and $\to v | | \to E$ , then $\to v$ must be parallel to $\to B .$
$∴_{L} = q \to E + 0 = q \to E$
Charge will either accelerate or decelerate in the same direction of $\to v,$ but does not deflect.

If $\to E$ is perpendicular to $\to B$ and $\to v$ is parallel to $\to E \times \to B$
$\Rightarrow$ $\to E,$ $\to B$ and $\to v$ are mutually perpendicular to each other.
$\Rightarrow \to v \times \to B = k \to E$

$_{L} = q \to E + q (\to v \times \to B)$
$_{L} = q \to E + q k \to E$
For $k = - 1, F_{L} = 0$
Hence, its possible for charge to travel undeflected.

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