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A magnetic field set up using Helmholtz coils (described in question 4.16) is uniform in a small region and has magnitude of $0.75 \mathrm{~T}$. 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 $9.0 \times 10^{-5} \mathrm{~V} \mathrm{~m}^{-1}$, make a simple guess as to what the beam contains. Why is the answer not unique?
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Verified Answer
Narrow beam of charged particles remains undeflected and is perpendicular to both electric field and magnetic fields which are mutually perpendicular. So, the electric force is balanced by magnetic force.
$q E=q v \mathrm{~B}$
speed of charged particles
$v=E / B=\frac{9 \times 10^{-5}}{0.75}=12 \times 10^{-5} \mathrm{~ms}^{-1}$
Because the beam is accelerated through $15 \mathrm{kV}$, if charge is $q$, then kinetic energy gained by charged particles
$\frac{1}{2} m v^2=q V \Rightarrow \frac{m}{q}=\frac{2 V}{v^2}$
$$
=\frac{2 \times 15 \times 10^3}{\left(12 \times 10^{-5}\right)^2}=20.8 \times 10^1
$$
Here, we can only obtain charge to mass ratio and same ratio can be in Deuterium ions, $\mathrm{He}^{++}, \mathrm{Li}^{++}$, so the beam can contain any of these charged particles.
$q E=q v \mathrm{~B}$
speed of charged particles
$v=E / B=\frac{9 \times 10^{-5}}{0.75}=12 \times 10^{-5} \mathrm{~ms}^{-1}$
Because the beam is accelerated through $15 \mathrm{kV}$, if charge is $q$, then kinetic energy gained by charged particles
$\frac{1}{2} m v^2=q V \Rightarrow \frac{m}{q}=\frac{2 V}{v^2}$
$$
=\frac{2 \times 15 \times 10^3}{\left(12 \times 10^{-5}\right)^2}=20.8 \times 10^1
$$
Here, we can only obtain charge to mass ratio and same ratio can be in Deuterium ions, $\mathrm{He}^{++}, \mathrm{Li}^{++}$, so the beam can contain any of these charged particles.
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