Charges in magnetic fields explained: HSC Physics Module 6

The magnetic force provides the centripetal force for circular motion:

$qvB = \frac{m v^2}{r}$

Solving for $r$:

$r = \frac{mv}{qB} = \frac{1.67 \times 10^{-27} \times 2.5 \times 10^6}{1.60 \times 10^{-19} \times 0.40}$
$r = \frac{4.18 \times 10^{-21}}{6.40 \times 10^{-20}} = 6.5 \times 10^{-2}$ m = 6.5 cm.

Period:

$T = \frac{2 \pi r}{v} = \frac{2 \pi m}{q B} = \frac{2 \pi \times 1.67 \times 10^{-27}}{1.60 \times 10^{-19} \times 0.40}$
$T = \frac{1.05 \times 10^{-26}}{6.40 \times 10^{-20}} = 1.6 \times 10^{-7}$ s.

Markers reward the explicit equating of magnetic and centripetal force, both formulas $r = mv/(qB)$ and $T = 2 \pi m / (qB)$, and final answers with units.

The magnetic force on a moving charge is $\vec{F} = q \vec{v} \times \vec{B}$. The cross product means $\vec{F}$ is always perpendicular to $\vec{v}$ and to $\vec{B}$.

Because $\vec{F}$ is perpendicular to $\vec{v}$, the magnetic force does no work on the particle ($W = \vec{F} \cdot \vec{d} = 0$). The kinetic energy and hence the speed of the particle remain constant.

A constant-magnitude force perpendicular to the velocity provides centripetal acceleration, which causes circular motion. If the particle has no velocity component along $\vec{B}$, the motion is a closed circle; if it does, the motion is a helix (which is outside the standard HSC treatment).

Markers reward the perpendicularity statement, the no-work argument for constant speed, and the centripetal-force conclusion.

The velocity is perpendicular to the field, so use $F = qvB$ (the $\sin\theta$ factor is $1$).

$F = qvB = (1.602 \times 10^{-19})(3.0 \times 10^6)(0.15)$
$F = 7.2 \times 10^{-14}\ \text{N}$.

Marks: one for the correct formula with values substituted, one for the answer stated to two significant figures with the unit newton.

(a) Radius. The magnetic force is the centripetal force, so $r = \dfrac{mv}{qB}$.

$r = \dfrac{(1.673 \times 10^{-27})(2.0 \times 10^6)}{(1.602 \times 10^{-19})(0.30)} = \dfrac{3.35 \times 10^{-21}}{4.81 \times 10^{-20}} = 7.0 \times 10^{-2}\ \text{m}$.

(b) Period. $T = \dfrac{2\pi m}{qB} = \dfrac{2\pi (1.673 \times 10^{-27})}{(1.602 \times 10^{-19})(0.30)} = 2.2 \times 10^{-7}\ \text{s}$.

Marks: one for the radius formula with substitution, one for $r = 7.0 \times 10^{-2}$ m, one for $T = 2.2 \times 10^{-7}$ s. Note $T$ does not use the speed - the period is independent of $v$.

(a) For no deflection the electric and magnetic forces must balance: $qE = qvB$. Cancelling $q$ gives the selected speed $v = \dfrac{E}{B}$, independent of charge and mass.

(b) $v = \dfrac{E}{B} = \dfrac{4.0 \times 10^3}{0.025} = 1.6 \times 10^5\ \text{m/s}$.

(c) For a faster ion the magnetic force $qvB$ exceeds the electric force $qE$, so there is a net force in the direction of the magnetic force and the ion is deflected out of the beam (it does not reach the exit slit).

Marks: one for the force balance giving $v = E/B$, one for the correct speed with unit, one for correctly reasoning that faster ions are over-deflected by the (now larger) magnetic force.

(a) Speed from the accelerating voltage. All the electrical work becomes kinetic energy: $qV = \tfrac{1}{2}mv^2$, so $v = \sqrt{\dfrac{2qV}{m}}$.

$v = \sqrt{\dfrac{2(1.602 \times 10^{-19})(800)}{3.986 \times 10^{-26}}} = \sqrt{6.43 \times 10^{9}} = 8.0 \times 10^4\ \text{m/s}$.

(b) Radius in the field. $r = \dfrac{mv}{qB} = \dfrac{(3.986 \times 10^{-26})(8.0 \times 10^4)}{(1.602 \times 10^{-19})(0.25)} = 8.0 \times 10^{-2}\ \text{m}$.

Marks: one for $qV = \tfrac{1}{2}mv^2$, one for $v = 8.0 \times 10^4$ m/s, one for $r = mv/(qB)$ with substitution, one for $r = 8.0 \times 10^{-2}$ m with the unit. This is the mass-spectrometer principle: heavier isotopes trace larger radii.

(a) The graph is a straight line through the origin, so the radius is directly proportional to the speed ($r \propto v$), as expected from $r = \dfrac{mv}{qB}$ with $m$, $q$ and $B$ constant.

(b) Gradient $= \dfrac{\Delta r}{\Delta v} = \dfrac{(20.9 - 5.2) \times 10^{-2}\ \text{m}}{(4.0 - 1.0) \times 10^{6}\ \text{m/s}} = \dfrac{0.157}{3.0 \times 10^{6}} = 5.2 \times 10^{-8}\ \text{s}$.

(c) The gradient equals $\dfrac{m}{qB}$, so $B = \dfrac{m}{q \times \text{gradient}} = \dfrac{1.673 \times 10^{-27}}{(1.602 \times 10^{-19})(5.2 \times 10^{-8})} = 0.20\ \text{T}$.

Marks: one for identifying direct proportionality ($r \propto v$, line through origin), one for a correct gradient with units of seconds, one for equating the gradient to $m/(qB)$, one for $B = 0.20$ T.

Band-6 plan. (1) State the role of the accelerating (electric) stage with $qV = \tfrac{1}{2}mv^2$. (2) State the role of the magnetic stage with $r = mv/(qB)$. (3) Combine them to show $r$ depends on $\sqrt{m}$. (4) Conclude that different masses land at different radii, achieving separation. Name each equation and link it to a physical effect.

Model answer. In the first stage a potential difference $V$ accelerates each singly charged ion from rest. The electric field does work $qV$ on the ion, all of which becomes kinetic energy, so $qV = \tfrac{1}{2}mv^2$ and the ion enters the field region with speed $v = \sqrt{2qV/m}$. Because $q$ and $V$ are the same for every ion, a heavier isotope leaves the accelerator moving more slowly.

In the second stage the ion moves perpendicular to a uniform magnetic field, which exerts a force $F = qvB$ perpendicular to the velocity. This force does no work (it is always perpendicular to $v$), so the speed stays constant while the ion is bent into a circular arc. Setting the magnetic force equal to the centripetal force, $qvB = \dfrac{mv^2}{r}$, gives $r = \dfrac{mv}{qB}$.

Substituting the entry speed, $r = \dfrac{m}{qB}\sqrt{\dfrac{2qV}{m}} = \dfrac{1}{B}\sqrt{\dfrac{2mV}{q}}$, so the radius is proportional to $\sqrt{m}$ (for fixed $q$, $V$ and $B$). Heavier isotopes therefore travel on larger-radius arcs and strike the detector at a different position from lighter isotopes, and this spatial separation is what identifies and measures the relative abundance of each isotope.

Marker's note: the top band names both equations, explains that the magnetic force does no work (so speed and hence energy are set by the electric stage, not the magnetic one), and derives the key result $r \propto \sqrt{m}$ to justify separation. A response that merely describes "the fields bend the ions" without the $r \propto \sqrt{m}$ link caps in the middle band.

Band-6 plan. Build a three-criterion comparison: (1) direction of the force, (2) effect on speed/kinetic energy (work done), (3) resulting path. Give the equation for each field, treat both fields on the same criteria, and finish with the unifying idea (electric fields can speed a charge up; magnetic fields only steer it).

Model answer. In a uniform electric field the force on a charge is $F = qE$, directed along the field (for a positive charge) regardless of how the charge is moving. Because this force generally has a component along the velocity, it does work on the charge, so the kinetic energy and speed change. A charge released in a uniform electric field moves like a projectile: if launched across the field it follows a parabolic path, accelerating in the field direction.

In a uniform magnetic field the force is $F = qvB\sin\theta$, directed perpendicular to both the velocity and the field. Because the force is always perpendicular to the velocity it does no work, so the speed and kinetic energy stay constant. A charge moving perpendicular to the field is bent into a circle of radius $r = \dfrac{mv}{qB}$ at constant speed (a helix if it also has a velocity component along the field).

The decisive difference is energy: an electric field can transfer energy to a charge and change its speed, whereas a magnetic field can only change the direction of motion. This is why particle accelerators use electric fields to increase a particle's energy and magnetic fields to steer and confine the beam.

Marker's note: the top band addresses every criterion for both fields (not one field in detail and the other in passing), correctly states that the magnetic force does no work while the electric force can, and reaches the synthesising judgement (electric = accelerate, magnetic = steer). Naming the parabolic versus circular paths earns the "resulting path" marks.