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An ionized gas contains both positive and negative ions. If it is subjected simultaneously to an electric field along the + x-direction and a magnetic field along the +z-direction, then
An ionized gas contains both positive and negative ions. If it is subjected simultaneously to an electric field along the + x-direction and a magnetic field along the +z-direction, then
Updated On: Jun 14, 2022
- positive ions deflect towards +y-direction and negative ions towards -y-direction
- all ions deflect towards +y-direction
- all ions deflect towards y-direction
- positive ions deflect towards -y-direction and negative ions towards -y-direction
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The Correct Option is C
Solution and Explanation
We can write $E=E .\widehat{i}$ and $B=B\widehat{k}$
Velocity of the particle will be along E direction. Therefore, we can write
$\hspace15mm v=Aq E\widehat{i}$
In E, B and v. A, E and 5 are positive constants while q can be positive or negative
Now, magnetic force on the particle will be
$F_m=q(v \times B)=q\{AqE\widehat{i}\}\times\{B\widehat{k}\}$
$\hspace15mm =q^2 AEB(\widehat{i}\times\widehat{k})$
$\hspace15mm F_m=q^2 AEB (-\widehat{j})$
Since, $F_m$is along negative y-axis, no matter what is the sign of charge' Therefore, all ions will deflect towards negative y-direction.
Velocity of the particle will be along E direction. Therefore, we can write
$\hspace15mm v=Aq E\widehat{i}$
In E, B and v. A, E and 5 are positive constants while q can be positive or negative
Now, magnetic force on the particle will be
$F_m=q(v \times B)=q\{AqE\widehat{i}\}\times\{B\widehat{k}\}$
$\hspace15mm =q^2 AEB(\widehat{i}\times\widehat{k})$
$\hspace15mm F_m=q^2 AEB (-\widehat{j})$
Since, $F_m$is along negative y-axis, no matter what is the sign of charge' Therefore, all ions will deflect towards negative y-direction.
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