Motion in a Magnetic Field

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1. 1.↑ back to the question

   Here $\vec{F} = -e(\vec{v}\times\vec{B}) = -e(v\hat{i} \times B(-\hat{k})) = -evB\hat{j}$ Thus, the force is along the $-y$ axis initially. Since the force is always perpendicular to velocity (magnetic force does no work and changes only direction, not speed), the electron follows a circular arc curving towards $-y$. Thus the trajectory will have a semicircular shape in the $x$–$(-y)$ plane, i.e. the electron curves downward (into $-y$) and re-emerges moving along $-x$.

2. 2.↑ back to the question

   In a uniform magnetic field, the frequency of a charge particle moving in a circle is given by $\nu = \dfrac{Bq}{2\pi m}$. For given $B$: $\nu \propto \dfrac{q}{m}$ $\therefore \dfrac{\nu_p}{\nu_e} = \dfrac{q_p}{q_e}\times\dfrac{m_e}{m_p}$ As $q_e = q_p$ but $m_e < m_p$, hence $\nu_p < \nu_e$. So, the electron moves in a circular path with higher frequency.

3. 3.↑ back to the question

   The magnetic force $\vec{F}_B = q(\vec{v}\times\vec{B})$ is always perpendicular to $\vec{v}$ and $\vec{B}$, and hence cannot change the speed of the particle. In other words, the magnetic force cannot speed up or slow down a charged particle. Hence, the kinetic energy of the particle remains unaffected (unchanged).

4. 4.↑ back to the question

   A charged particle moving in a uniform magnetic field has two motion components: (i) a linear motion along the direction of $\vec{B}$ (since the component of velocity parallel to $B$ experiences no force), and (ii) a circular motion in the plane perpendicular to $\vec{B}$ (since the component of velocity perpendicular to $B$ is continuously deflected). The resultant path of the particle is thus a helix, with its axis along the direction of $\vec{B}$. The velocity component along $B$ is $v_\parallel = v\cos30°$, and the component perpendicular to $B$ is $v_\perp = v\sin30°$; the perpendicular component executes circular motion of radius $r = \dfrac{mv_\perp}{qB} = \dfrac{mv\sin30°}{qB}$, while the parallel component carries the particle steadily along $B$, together tracing a helical path.

5. 5.↑ back to the question

   When a proton, deuteron and alpha particle are each accelerated through the same potential difference $V$, their kinetic energies are: $E_p = eV$, $E_d = eV$ (deuteron charge $= e$), $E_\alpha = 2eV$ (alpha particle charge $= 2e$) (i) $KE_p : KE_d : KE_\alpha = eV : eV : 2eV = 1 : 1 : 2$ (ii) Radius in magnetic field: $r = \dfrac{mv}{qB} = \dfrac{\sqrt{2mKE}}{qB}$ $r_p : r_d : r_\alpha = \dfrac{\sqrt{m_p \cdot eV}}{e} : \dfrac{\sqrt{m_d \cdot eV}}{e} : \dfrac{\sqrt{m_\alpha \cdot 2eV}}{2e}$ With $m_d = 2m_p$, $m_\alpha = 4m_p$: $r_p : r_d : r_\alpha = \dfrac{\sqrt{m_p}}{e} : \dfrac{\sqrt{2m_p}}{e} : \dfrac{\sqrt{4m_p}}{2e} = 1 : \sqrt{2} : 1$ Given $r_p = 5$ cm, so $r_\alpha = 5$ cm (equal to $r_p$) and $r_d = 5\sqrt{2}$ cm.

6. 6.↑ back to the question

   (a) The magnetic force on the moving charge is $\vec{F} = q(\vec{v}\times\vec{B})$. This force acts perpendicular to the velocity vector at all instants; it therefore only changes the direction of velocity, not its magnitude. Since the magnitude of speed remains constant while the direction of the force is always perpendicular to velocity (i.e. towards a fixed centre), the particle moves in a circular path, with the magnetic force providing the necessary centripetal force. So, $F = qvB\sin\theta$ with $\theta = 90°$: $F = qvB$ Equating to the centripetal force: $\dfrac{mv^2}{r} = qvB \Rightarrow r = \dfrac{mv}{qB}$ (b) If $\theta \ne 90°$, the velocity has a component along $\vec{B}$ (which experiences no force) and a component perpendicular to $\vec{B}$ (which executes circular motion). The particle then moves along $B$ with uniform velocity while circling around it, so its motion becomes helical.

7. 7.↑ back to the question

   (a) $\vec{F} = q(\vec{v}\times\vec{B})$ (i) The particle moves in a circular path when $\vec{v}$ is perpendicular to $\vec{B}$ (i.e. the entire velocity provides for circular motion, with no component along $B$). (ii) The particle moves in a helical path when $\vec{v}$ makes an angle $\theta$ ($\ne 90°$, $\ne 0°$) with $\vec{B}$ — the component of $\vec{v}$ along $\vec{B}$ produces uniform linear motion while the perpendicular component produces circular motion, and together they trace a helix. (b) The magnetic force $\vec{F} = q(\vec{v}\times\vec{B})$ always acts perpendicular to $\vec{v}$, so $\vec{F}\cdot\vec{v} = q(\vec{v}\times\vec{B})\cdot\vec{v} = 0$. By Newton's second law, $\vec{F} = m\dfrac{d\vec{v}}{dt}$, so $m\dfrac{d\vec{v}}{dt}\cdot\vec{v} = 0$, i.e. $\dfrac{m}{2}\left[\dfrac{d\vec{v}}{dt}\cdot\vec{v} + \vec{v}\cdot\dfrac{d\vec{v}}{dt}\right] = 0$ $\Rightarrow \dfrac{m}{2}\dfrac{d}{dt}(\vec{v}\cdot\vec{v}) = 0 \Rightarrow \dfrac{d}{dt}\left(\dfrac{1}{2}mv^2\right) = 0 \Rightarrow \dfrac{dK}{dt} = 0$ Hence $K =$ constant, i.e. the kinetic energy remains constant.

8. 8.↑ back to the question

   (a) Magnetic force acting on a charged particle $q$ moving with velocity $v$ in a uniform magnetic field $B$ is given by $\vec{F} = q(\vec{v}\times\vec{B})$. (b) Magnetic force on neutron, $F = 0$ (as its charge $q = 0$), so the neutron moves undeflected in a straight line. Magnetic force on the alpha particle, $\vec{F}_\alpha = q\vec{v}\times\vec{B} = 2evB$ (upward, since it is positively charged) — so its path curves upward, i.e. the force is perpendicular to the velocity and bends the path towards the upper side. Magnetic force on the electron, $\vec{F}_e = q\vec{v}\times\vec{B} = |-evB|$ (downward, since it is negatively charged) — so its path curves downward, opposite to the alpha particle's curve. For a charged particle moving perpendicular to $B$: $qvB = \dfrac{mv^2}{r} \Rightarrow r = \dfrac{mv}{qB}$ Since all three move with the same speed $v = v_\alpha = v_n = v_e$: radius of the alpha particle's path, $r_\alpha = \dfrac{m_\alpha v}{q_\alpha B} = \dfrac{4m_e v}{2eB} = \dfrac{2m_e v}{eB}$ (taking $m_\alpha \approx 4m_p \approx 4 \times 1836\,m_e$, but relative to the electron's own mass, $r_e = \dfrac{m_e v}{eB}$). So: the neutron goes straight through undeflected; the alpha particle curves in a large circular arc upward; the electron curves in a much smaller circular arc downward (since $m_e \ll m_\alpha$, giving $r_e \ll r_\alpha$).

9. 9.↑ back to the question

   Magnetic force on the charged particle: $\vec{F} = q(\vec{v}\times\vec{B})$; thus $|\vec{F}| = qvB\sin\theta$. Since the velocity has a component along $B$ and a component perpendicular to $B$: the parallel component ($v\cos\theta$) experiences no force (as $\sin0° = 0$) and moves the particle steadily along the $x$-axis, while the perpendicular component ($v\sin\theta$) is continuously deflected and executes uniform circular motion in the Y-Z plane. Thus radius of circular path is $r = \dfrac{mv\sin\theta}{qB}$. So the trajectory followed is a helical path, with axis along the direction of $\vec{B}$ (the $x$-axis). Time period of one rotation (circular component): $T = \dfrac{2\pi r}{v\sin\theta} = \dfrac{2\pi m}{qB}$ Horizontal distance moved by the particle in one rotation (pitch) $=$ (parallel velocity) $\times$ (time period): Pitch $= v\cos\theta \times T = \dfrac{2\pi m}{qB}v\cos\theta$

10. 10.↑ back to the question

    Let the time taken by the electron to come out of the region of magnetic field be $t$. Velocity of the electron, $v = 4\times10^4$ m/s; magnetic field, $B = 10^{-5}$ T; mass of the electron, $m = 9\times10^{-31}$ kg. The electron traces a circular (semicircular) arc within the field region and re-emerges from the same boundary it entered, so the time taken is that for a half-circle: $t = \dfrac{\pi r}{v}$, where $r = \dfrac{mv}{qB}$ $\Rightarrow t = \dfrac{\pi m}{Bq} = \dfrac{3.14\times9\times10^{-31}}{10^{-5}\times1.6\times10^{-19}}$ $t = 17.66\times10^{-7}$ s $= 1.77\ \mu$s Thus, the time taken by the electron to come out of the region of magnetic field is 1.77 μs, and its trajectory is a semicircular arc bending the electron back the way it came, curving in the plane perpendicular to $B$.

11. 11.↑ back to the question

    The charged particle moves in a circular path with a constant speed and is acted upon only by the magnetic field. The radius of the circular path is given by $qvB = \dfrac{mv^2}{r}$, i.e. $r = \dfrac{mv}{qB}$. Period of revolution, $T = \dfrac{2\pi r}{v} = \dfrac{2\pi}{v}\cdot\dfrac{mv}{qB} = \dfrac{2\pi m}{qB}$ Frequency of revolution, $f = \dfrac{1}{T} = \dfrac{qB}{2\pi m}$ Clearly, the frequency $f$ is independent of both $v$ and $r$ (they cancelled out of the expression), and since it depends only on $q$, $B$ and $m$, it is also independent of the energy of the particle.

12. 12.↑ back to the question

    Given: $\Delta V = 10\times10^3$ V $= 10^4$ V, $B = 2\times10^{-3}$ T. Since the charge is accelerated through $\Delta V$: $\dfrac{1}{2}mv^2 = q\Delta V \Rightarrow v = \sqrt{\dfrac{2q\Delta V}{m}}$ ...(i) Now when it enters the magnetic field (perpendicular to it, since $v$ is along $x$ and $B$ is along $y$): $qvB = \dfrac{mv^2}{r}$ $\Rightarrow r = \dfrac{mv}{qB} = \dfrac{m}{qB}\sqrt{\dfrac{2q\Delta V}{m}} = \sqrt{\dfrac{2m\Delta V}{B^2 q}}$ For the alpha particle: $q = 2e = 3.2\times10^{-19}$ C, $m = 6.4\times10^{-27}$ kg. $r = \sqrt{\dfrac{2\times6.4\times10^{-27}\times10^4}{(2\times10^{-3})^2\times2\times1.6\times10^{-19}}}$ $r = 10$ m

13. 13.↑ back to the question

    (i) Radius, $r = \dfrac{mv}{Bq} = \dfrac{m}{Bq}\sqrt{\dfrac{2K}{m}} = \dfrac{\sqrt{2mK}}{Bq}$ $m_\alpha = 4m$, $m_d = 2m$, $m_p = m$; $q_\alpha = 2e$, $q_d = e$, $q_p = e$. Here, KE is the same and magnetic field is constant, so $r \propto \dfrac{\sqrt{m}}{q}$ $r_\alpha \propto \dfrac{\sqrt{4m}}{2e} \propto \dfrac{\sqrt{m}}{e}$, $\quad r_d \propto \dfrac{\sqrt{2m}}{e}$, $\quad r_p \propto \dfrac{\sqrt{m}}{e}$ So, $r_\alpha : r_d : r_p = 1 : \sqrt{2} : 1$ (ii) The current flows $A\to C\to D\to E\to A$: rightward ($+x$) from $A$ to $C$, downward ($-y$) from $C$ to $D$, leftward ($-x$) from $D$ to $E$, and upward ($+y$) from $E$ back to $A$ — this is a clockwise circulation as viewed from the $+z$ axis (looking at the X-Y plane from outside the page). By the right-hand rule, a clockwise current (as seen from $+z$) gives a magnetic dipole moment $\vec{M}$ directed along $-z$ (into the page). For the coil to be in unstable equilibrium, its potential energy $U = -\vec{M}\cdot\vec{B}$ must be maximum, i.e. $\vec{M}$ and $\vec{B}$ must be anti-parallel ($\theta = 180°$). Since $\vec{M}$ is along $-z$, the magnetic field $\vec{B}$ must be directed along $+z$ (out of the page, along the positive $Z$-axis) for the coil to be in unstable equilibrium. (iii) If the resistance of the ammeter were high, the total resistance of the circuit would be high, due to which the current would decrease. So, to avoid change of current in the circuit being measured, we use a low resistance ammeter.

14. 14.Answer: (B)↑ back to the question

    Here the electrostatic (Coulomb) attraction between the two opposite charges $q$ and $-q$ provides the centripetal force for the circular motion. $F_e = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q(-q)}{r^2}$, magnitude $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{r^2}$ (attractive, since opposite charges) Centripetal force, $F_c = \dfrac{mv^2}{r}$ Equating: $\dfrac{mv^2}{r} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{r^2}$ $\Rightarrow r = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{mv^2} = \dfrac{1}{4\pi\varepsilon_0 m}\left(\dfrac{q}{v}\right)^2$

15. 15.Answer: (B)↑ back to the question

    Answer: (b) $\sqrt{2}\,r$ For a charged particle moving in a magnetic field: $qvB = \dfrac{mv^2}{r} \Rightarrow r = \dfrac{mv}{qB}$ ...(i) Since it is accelerated through potential difference $V$: $qV = \dfrac{1}{2}mv^2 \Rightarrow v = \sqrt{\dfrac{2qV}{m}}$ ...(ii) Substituting (ii) in (i): $r = \dfrac{m}{qB}\sqrt{\dfrac{2qV}{m}} = \dfrac{1}{B}\sqrt{\dfrac{2mV}{q}}$ ...(iii) Now with $V' = 2V$: $r' = \dfrac{1}{B}\sqrt{\dfrac{2m(2V)}{q}} = \sqrt{2}\cdot\dfrac{1}{B}\sqrt{\dfrac{2mV}{q}} = \sqrt{2}\,r$

16. 16.Answer: (C)↑ back to the question

    Answer: (c) $\left(\dfrac{r_1}{r_2}\right)^2$ Radius, $r = \dfrac{mv}{Bq} = \dfrac{m}{Bq}\sqrt{\dfrac{2K}{m}} = \dfrac{\sqrt{2mK}}{Bq}$, where $K = qV$ is the kinetic energy gained. $\therefore r \propto \sqrt{m}$, i.e. $m \propto r^2$ (since $B$, $q$ and $K \propto q$ are the same for both particles as they carry equal charge and are accelerated through the same $V$). So, $\dfrac{m_1}{m_2} = \left(\dfrac{r_1}{r_2}\right)^2$