Question

An electron (mass$=9.1\times {{10}^{-31}}kg;$charge $=1.6\times {{10}^{-19}}C$) experiences no deflection if subjected to an electric field of $3.2\times {{10}^{5}}v/m$, and a magnetic field of $2.0\times {{10}^{-3}}Wb/{{m}^{2}}$. Both the fields are normal to the path of electrons and each other. If the electric field is removed, then the electron will revolve in an orbit of radius
A.$45m$
B.$4.5m$
C.$0.45m$
D.$0.045m$

Answer 1

Hint: When an electron enters both electric and magnetic fields, then the magnetic force exerted on a charged particle moving through both magnetic and electric fields is called Lorentz force. Here Lorentz force is zero due to no deflection. Then we can calculate the velocity of the electron and then we can easily calculate the radius of the circular orbit by removing the electric field

Formula used:
The velocity $v$ of a charged particle (No deflection) is given by:
$v=\dfrac{E}{B}$
Here $E\And B$ is the electric and magnetic field respectively.
Radius,$r=\dfrac{mv}{qB}$
Where $m\And q$ denotes the mass and charge of the particle.


Complete answer:

The magnetic force $F$ which is exerted on a charged particle $q$ like an electron moving through both a magnetic field,$B$ and an electric field, $E$ is known as Lorentz force.
The electrical force${{F}_{Electric}}$ is generated due to the presence of an electric field,$E$ then
${{F}_{Electric}}=q\times E$
Here $q$is the charge of an electron.
The magnetic force ${{F}_{Magnetic}}$ acts on an electron moving with uniform velocity $\vec{v}$ through the magnetic field $\vec{B}$ then
${{\vec{F}}_{Magnetic}}=q(\vec{B}\times \vec{v})$
Or,${{F}_{Magnetic}}=qBv\sin {{90}^{o}}$ [Since the angle between $\vec{v}\And \vec{B}$ is ${{90}^{o}}$as they are mutually perpendicular to each other]
Or,${{F}_{Magnetic}}=qvB$
 Since there is no deflection i.e. $F=0$ the magnetic and electric fields are equal and opposite in magnitude.
$\therefore qE=qvB$
Or,$v=\dfrac{E}{B}$ ……(i)
Given, $E=3.2\times {{10}^{5}}v/m$and $B=2.0\times {{10}^{-3}}wb/{{m}^{2}}$
$v=\dfrac{3.2\times {{10}^{5}}v/m}{2.0\times {{10}^{-3}}wb/{{m}^{2}}}=1.6\times {{10}^{8}}m/s$
As the electric field is removed, the electron will move in a circular orbit due to the perpendicular magnetic field.
Then the radius of a circular orbit, $r=\dfrac{mv}{qB}$ ……..(ii)
Given the mass of the electron, $m=9.1\times {{10}^{-31}}kg$
Charge of the electron, $q=1.6\times {{10}^{-19}}C$
Putting the value of $v$in equation (ii),
$\therefore r=\dfrac{9.1\times {{10}^{-31}}kg\times 1.6\times {{10}^{8}}m/s}{1.6\times {{10}^{-19}}C\times 2\times {{10}^{-3}}wb/{{m}^{2}}}=0.455m$
Therefore electrons will revolve in an orbit of radius $0.455m$.

Thus, option (C) is correct.




Note: Lorentz force will be zero under certain conditions. One, if the particle has no charge, and second when the velocity of the charged particle is parallel to the magnetic field. And it will be maximum when the charge moves perpendicular to the magnetic field.