
An electron enters a region where the electrostatic field is $20N/C$and magnetic field is $5T$. If an electron passes undeflected through the region, then the velocity of the electron will be
A.$0.25m{{s}^{-1}}$ B.$2m{{s}^{-1}}$
C.$4m{{s}^{-1}}$
D.$8m{{s}^{-1}}$
Answer
162.3k+ views
Hint: When an electron with charge perpendicular to the electric and magnetic field in a region of the crossed field, electrons with velocity do not deviate. Therefore it must satisfy the condition that magnetic force and electrostatic force are equal. In this way, we get the velocity formula by taking these two forces (magnetic & electric forces) equal.
Formula used:
If an electron passes undeflected through the region, then velocity$v$is given by,
$v=\dfrac{E}{B}$
Here $v=$velocity of the electron
$E=$electrostatic field
$B=$magnetic field
Complete answer:
According to the definition of Lorentz force, the combined forces on moving charges due to the electromagnetic force i.e, both electric and magnetic forces.
Let an electron with charge$e$, moves with velocity $\vec{v}$, in the region of both electric and magnetic fields. Then the force (${{\vec{F}}_{E}}$) on the electron due to electric field ($\vec{E}$) is given by,
${{\vec{F}}_{E}}=e\times \vec{E}$ …….(i)
And the force (${{\vec{F}}_{B}}$) due to the magnetic field, ($\vec{B}$) is given by,
${{\vec{F}}_{B}}=e(\vec{v}\times \vec{B})$ ………(ii)
Here $\vec{v}=$velocity of the electron
Now if the electron enters a region of the crossed field, where it is perpendicular to the electric and magnetic field, the velocity of that electron does not deviate.
In this situation ${{\vec{F}}_{E}}={{\vec{F}}_{B}}$
Or,$e\times \vec{E}=e(\vec{v}\times \vec{B})$
Or,$\vec{E}=(\vec{v}\times \vec{B})$
Or,$E=vB\sin \theta $ [since $\theta ={{90}^{o}}$=angle between $\vec{B}\And \vec{v}$]
Hence,$E=vB\sin {{90}^{o}}$
Or,$E=vB$
Or,$v=\dfrac{E}{B}$
Now putting the given values, $E=20N/C$ and $B=5T$
$\therefore v=\dfrac{20N/C}{5T}=4m{{s}^{-1}}$
Therefore, the velocity of the electron will be $4m{{s}^{-1}}$.
Thus, option (C) is correct.
Note:If an electron is in the region where the electric field, magnetic field, and direction of velocity are parallel to each other, then the force due to the magnetic field is zero. The charged particle will be accelerated by the electric field only.
Formula used:
If an electron passes undeflected through the region, then velocity$v$is given by,
$v=\dfrac{E}{B}$
Here $v=$velocity of the electron
$E=$electrostatic field
$B=$magnetic field
Complete answer:
According to the definition of Lorentz force, the combined forces on moving charges due to the electromagnetic force i.e, both electric and magnetic forces.
Let an electron with charge$e$, moves with velocity $\vec{v}$, in the region of both electric and magnetic fields. Then the force (${{\vec{F}}_{E}}$) on the electron due to electric field ($\vec{E}$) is given by,
${{\vec{F}}_{E}}=e\times \vec{E}$ …….(i)
And the force (${{\vec{F}}_{B}}$) due to the magnetic field, ($\vec{B}$) is given by,
${{\vec{F}}_{B}}=e(\vec{v}\times \vec{B})$ ………(ii)
Here $\vec{v}=$velocity of the electron
Now if the electron enters a region of the crossed field, where it is perpendicular to the electric and magnetic field, the velocity of that electron does not deviate.
In this situation ${{\vec{F}}_{E}}={{\vec{F}}_{B}}$
Or,$e\times \vec{E}=e(\vec{v}\times \vec{B})$
Or,$\vec{E}=(\vec{v}\times \vec{B})$
Or,$E=vB\sin \theta $ [since $\theta ={{90}^{o}}$=angle between $\vec{B}\And \vec{v}$]
Hence,$E=vB\sin {{90}^{o}}$
Or,$E=vB$
Or,$v=\dfrac{E}{B}$
Now putting the given values, $E=20N/C$ and $B=5T$
$\therefore v=\dfrac{20N/C}{5T}=4m{{s}^{-1}}$
Therefore, the velocity of the electron will be $4m{{s}^{-1}}$.
Thus, option (C) is correct.
Note:If an electron is in the region where the electric field, magnetic field, and direction of velocity are parallel to each other, then the force due to the magnetic field is zero. The charged particle will be accelerated by the electric field only.
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