A magnetic field set up using Helmholtz coils is uniform in a small region and has a magnitude of $0.75 \;T$. In the same region, a uniform electrostatic field is maintained in a direction normal to the common axis of the coils. A narrow beam of (single species) charged particles all accelerated through $15\; kV$ enters this region in a direction perpendicular to both the axis of the coils and the electrostatic field. If the beam remains undeflected when the electrostatic field is $9.0 \times 10^{-5} \;V\, m ^{-1},$ make a simple guess as to what the beam contains. Why is the answer not unique?
A magnetic field set up using Helmholtz coils is uniform in a small region and has a magnitude of $0.75 \;T$. In the same region, a uniform electrostatic field is maintained in a direction normal to the common axis of the coils. A narrow beam of (single species) charged particles all accelerated through $15\; kV$ enters this region in a direction perpendicular to both the axis of the coils and the electrostatic field. If the beam remains undeflected when the electrostatic field is $9.0 \times 10^{-5} \;V\, m ^{-1},$ make a simple guess as to what the beam contains. Why is the answer not unique?
Magnetic field, $B=0.75 \,T$ Accelerating voltage, $V =15\, kV =15 \times 10^{3} \,V$
Electrostatic field, $E=9 \times 10^{5} \,Vm ^{-1}$
Mass of the electron $=m$ Charge of the electron $=e$ Velocity of the electron $=v$ Kinetic energy of the electron $=e V$ $\Rightarrow \frac{1}{2} m v^{2}=e V$
$\therefore \frac{e}{m}=\frac{v^{2}}{2 V}\dots(i)$
since the particle remains undeflected by electric and magnetic fields, we can infer that the electric field is balancing the magnetic field.
$\therefore e E=e v B$
$v=\frac{E}{B}\ldots(ii)$
Putting equation $(ii)$ in equation $(i),$ we get
$\frac{e}{m}=\frac{1}{2} \frac{\left(\frac{E}{B}\right)^{2}}{V}=\frac{E^{2}}{2 V B^{2}}$
$=\frac{\left(9.0 \times 10^{5}\right)^{2}}{2 \times 15000 \times(0.75)^{2}}=4.8 \times 10^{7} \,C / kg$
This value of specific charge e/m is equal to the value of deuteron or deuterium ions. This is not a unique answer. Other possible answers are $H e^{++}, L i^{+++}$
Similar Questions
When a charged particle moving with velocity $\vec v$ is subjected to a magnetic field of induction $\vec B$, the force on it is non-zero. This implies that
When a charged particle moving with velocity $\vec v$ is subjected to a magnetic field of induction $\vec B$, the force on it is non-zero. This implies that
- [AIPMT 2006]
A particle of mass $m = 1.67 \times 10^{-27}\, kg$ and charge $q = 1.6 \times 10^{-19} \, C$ enters a region of uniform magnetic field of strength $1$ $tesla$ along the direction shown in the figure. If the direction of the magnetic field is along the outward normal to the plane of the paper, then the time spent by the particle in the region of the magnetic field after entering it at $C$ is nearly :-......$ns$
A particle of mass $m = 1.67 \times 10^{-27}\, kg$ and charge $q = 1.6 \times 10^{-19} \, C$ enters a region of uniform magnetic field of strength $1$ $tesla$ along the direction shown in the figure. If the direction of the magnetic field is along the outward normal to the plane of the paper, then the time spent by the particle in the region of the magnetic field after entering it at $C$ is nearly :-......$ns$
A positive charge $'q'$ of mass $'m'$ is moving along the $+ x$ axis. We wish to apply a uniform magnetic field $B$ for time $\Delta t$ so that the charge reverses its direction crossing the $y$ axis at a distance $d.$ Then
A positive charge $'q'$ of mass $'m'$ is moving along the $+ x$ axis. We wish to apply a uniform magnetic field $B$ for time $\Delta t$ so that the charge reverses its direction crossing the $y$ axis at a distance $d.$ Then
- [JEE MAIN 2014]
An electron is moving along the positive $x$-axis. If the uniform magnetic field is applied parallel to the negative $z$-axis. then
$A.$ The electron will experience magnetic force along positive $y$-axis
$B.$ The electron will experience magnetic force along negative $y$-axis
$C.$ The electron will not experience any force in magnetic field
$D.$ The electron will continue to move along the positive $x$-axis
$E.$ The electron will move along circular path in magnetic field
Choose the correct answer from the options given below:
An electron is moving along the positive $x$-axis. If the uniform magnetic field is applied parallel to the negative $z$-axis. then
$A.$ The electron will experience magnetic force along positive $y$-axis
$B.$ The electron will experience magnetic force along negative $y$-axis
$C.$ The electron will not experience any force in magnetic field
$D.$ The electron will continue to move along the positive $x$-axis
$E.$ The electron will move along circular path in magnetic field
Choose the correct answer from the options given below:
- [JEE MAIN 2023]
The radius of circular path of an electron when subjected to a perpendicular magnetic field is
The radius of circular path of an electron when subjected to a perpendicular magnetic field is