Fera Academy

Paper 2 Answers

Time2 hours

Marks60

SetSet 2

PaperLevel 1 - Physics Paper 2

Information

Section A: Electromagnetism
Section B: Relativity and Quantum Mechanics

Marking Guidance

These worked answers show the expected method and final result. Equivalent correct reasoning should receive credit.

Constants

Gravitational acceleration	\( g=9.81\,\mathrm{m\,s^{-2}} \)
Speed of light	\( c=3.00\times10^8\,\mathrm{m\,s^{-1}} \)
Elementary charge	\( e=1.60\times10^{-19}\,\mathrm{C} \)
Electron mass	\( m_e=9.11\times10^{-31}\,\mathrm{kg} \)
Planck constant	\( h=6.63\times10^{-34}\,\mathrm{J\,s} \)
Permittivity of free space	\( \epsilon_0=8.85\times10^{-12}\,\mathrm{F\,m^{-1}} \)
Magnetic constant	\( \mu_0=1.26\times10^{-6}\,\mathrm{N\,A^{-2}} \)
Boltzmann constant	\( k_B=1.38\times10^{-23}\,\mathrm{J\,K^{-1}} \)

Section A: Electromagnetism

Answer 1. Hall probe on a rotating field mapper

[15 marks]

A Hall probe is fixed to the end of a rotating field-mapper arm. At one instant the sensor strip has its length along \(+\hat{\imath}\), its width \(w=1.20\,\mathrm{mm}\) along \(+\hat{\jmath}\), and its thickness \(t=0.200\,\mathrm{mm}\) along \(+\hat{k}\). A steady conventional current \(I=18.0\,\mathrm{mA}\) flows along \(+\hat{\imath}\). The local magnetic field to be mapped is \(\vec B=0.420\,\mathrm{T}\,\hat{k}\). The mobile carriers in the semiconductor are electrons with density \(n=3.20\times10^{22}\,\mathrm{m^{-3}}\).

a) Find the electron drift velocity in vector form.

[4 marks]

The cross-sectional area normal to the current is \[ A=wt=(1.20\times10^{-3})(0.200\times10^{-3})=2.40\times10^{-7}\,\mathrm{m^2}. \] For electrons, \(q=-e\) and \[ \vec J=nq\vec v_d. \] The conventional current density is \(\vec J=(I/A)\hat{\imath}\), so \[ \vec v_d=\frac{\vec J}{nq} =-\frac{I}{neA}\hat{\imath}. \] Numerically, \[ |\vec v_d|=\frac{0.0180}{(3.20\times10^{22})(1.60\times10^{-19})(2.40\times10^{-7})} =14.6\,\mathrm{m\,s^{-1}}. \] Thus \[ \vec v_d=-14.6\,\mathrm{m\,s^{-1}}\,\hat{\imath}. \] The electron drift is opposite the conventional current.

Answer 1. Hall probe on a rotating field mapper

[15 marks]

b) Calculate the Hall electric field and state which side of the strip becomes negative.

[4 marks]

In steady state the sideways force on a carrier is zero: \[ q(\vec E_H+\vec v_d\times\vec B)=0. \] Therefore \[ \vec E_H=-\vec v_d\times\vec B. \] Here \[ \vec v_d\times\vec B=(-14.6\hat{\imath})\times(0.420\hat{k}) =+6.15\hat{\jmath}\,\mathrm{V\,m^{-1}}, \] so \[ \vec E_H=-6.15\,\mathrm{V\,m^{-1}}\,\hat{\jmath}. \] The magnetic force on each electron is \[ \vec F_B=q\vec v_d\times\vec B =(-e)(6.15\hat{\jmath}) =-9.84\times10^{-19}\,\mathrm{N}\,\hat{\jmath}. \] Electrons are pushed toward the \(-y\) side, so that side becomes negative relative to the \(+y\) side.

Answer 1. Hall probe on a rotating field mapper

[15 marks]

c) Derive the Hall voltage magnitude and calculate its value.

[3 marks]

The Hall voltage magnitude across the width is \[ |V_H|=E_Hw=v_dBw. \] Using \(v_d=I/(newt)\), \[ |V_H|=\frac{IB}{net}. \] For the given data, \[ |V_H|=\frac{(0.0180)(0.420)}{(3.20\times10^{22})(1.60\times10^{-19})(0.200\times10^{-3})} =7.38\times10^{-3}\,\mathrm{V}. \] Thus the expected Hall reading is \(7.38\,\mathrm{mV}\) in magnitude, with the sign set by the negative charge on the \(-y\) side.

Answer 1. Hall probe on a rotating field mapper

[15 marks]

d) A calibration run gives \(|V_H|=7.50\,\mathrm{mV}\). Infer the carrier density from this reading.

[2 marks]

Rearrange the Hall-voltage expression: \[ n=\frac{IB}{e t |V_H|}. \] If the measured magnitude is \(7.50\,\mathrm{mV}\), \[ n=\frac{(0.0180)(0.420)}{(1.60\times10^{-19})(0.200\times10^{-3})(7.50\times10^{-3})} =3.15\times10^{22}\,\mathrm{m^{-3}}. \] This is close to the stated value, so the reading is consistent with the carrier density.

Answer 1. Hall probe on a rotating field mapper

[15 marks]

e) Estimate the magnetic field produced by the probe current \(25.0\,\mathrm{mm}\) from the current path, and compare it with the field being mapped.

[2 marks]

Approximate the sensor current path as a long straight wire. At distance \(r=25.0\,\mathrm{mm}\), \[ B_{\mathrm{self}}=\frac{\mu_0 I}{2\pi r}. \] Thus \[ B_{\mathrm{self}}=\frac{(1.26\times10^{-6})(0.0180)}{2\pi(25.0\times10^{-3})} =1.44\times10^{-7}\,\mathrm{T}. \] Compared with the mapped field, \[ \frac{B_{\mathrm{self}}}{0.420}=3.43\times10^{-7}. \] The probe current's own field is negligible for this measurement scale.

Answer 2. Charging and dumping a pulsed electromagnet

[15 marks]

A laboratory pulsed electromagnet is modelled as a coil with inductance \(L=0.240\,\mathrm{H}\) and resistance \(R=3.00\,\Omega\). At \(t=0\), a \(24.0\,\mathrm{V}\) dc supply is connected in series with the coil. Later, the supply is disconnected and the coil is connected across a \(12.0\,\Omega\) dump resistor, so the discharge resistance is \(15.0\,\Omega\) including the coil.

a) Write the current-growth equation, then calculate the time constant and final current.

[3 marks]

For current growth in a series \(RL\) circuit, Kirchhoff's loop law gives \[ \mathcal E-IR-L\frac{dI}{dt}=0. \] The time constant is \[ \tau=\frac{L}{R}=\frac{0.240}{3.00}=8.00\times10^{-2}\,\mathrm{s}. \] After a long time \(dI/dt=0\), so the final current is \[ I_f=\frac{\mathcal E}{R}=\frac{24.0}{3.00}=8.00\,\mathrm{A}. \]

Answer 2. Charging and dumping a pulsed electromagnet

[15 marks]

b) Find the current \(0.120\,\mathrm{s}\) after the supply is connected.

[3 marks]

The growth solution is \[ I(t)=I_f\left(1-e^{-t/\tau}\right). \] At \(t=0.120\,\mathrm{s}\), \(t/\tau=0.120/0.0800=1.50\). Hence \[ I=8.00(1-e^{-1.50}) =8.00(0.777) =6.21\,\mathrm{A}. \] The coil has reached about \(78\%\) of its final current.

Answer 2. Charging and dumping a pulsed electromagnet

[15 marks]

c) At the same instant, calculate \(dI/dt\) and the self-induced emf of the inductor, including its sign relative to the current rise.

[4 marks]

During current growth, \[ \frac{dI}{dt}=\frac{\mathcal E}{L}e^{-t/\tau}. \] At \(t=0.120\,\mathrm{s}\), \[ \frac{dI}{dt}=\frac{24.0}{0.240}e^{-1.50} =22.3\,\mathrm{A\,s^{-1}}. \] Using the passive sign convention, the inductor voltage is \[ V_L=L\frac{dI}{dt}=(0.240)(22.3)=5.36\,\mathrm{V}. \] The self-induced emf is opposite to the increase in current, so \[ \mathcal E_L=-L\frac{dI}{dt}=-5.36\,\mathrm{V} \] for the chosen current direction.

Answer 2. Charging and dumping a pulsed electromagnet

[15 marks]

d) Find the magnetic energy stored in the coil at \(t=0.120\,\mathrm{s}\).

[2 marks]

The magnetic energy stored in the coil is \[ U_B=\frac12LI^2. \] At \(I=6.21\,\mathrm{A}\), \[ U_B=\frac12(0.240)(6.21)^2 =4.63\,\mathrm{J}. \]

Answer 2. Charging and dumping a pulsed electromagnet

[15 marks]

e) If the dump circuit is connected when the current is \(6.21\,\mathrm{A}\), find the time for the current to fall to \(0.500\,\mathrm{A}\).

[3 marks]

During discharge the total resistance is \[ R_{\mathrm{dis}}=3.00+12.0=15.0\,\Omega. \] The discharge time constant is \[ \tau_{\mathrm{dis}}=\frac{L}{R_{\mathrm{dis}}}=\frac{0.240}{15.0}=1.60\times10^{-2}\,\mathrm{s}. \] The decay is \[ I(t)=I_0e^{-t/\tau_{\mathrm{dis}}}. \] With \(I_0=6.21\,\mathrm{A}\) and \(I=0.500\,\mathrm{A}\), \[ 0.500=6.21e^{-t/\tau_{\mathrm{dis}}}, \] so \[ t=\tau_{\mathrm{dis}}\ln\left(\frac{6.21}{0.500}\right) =(1.60\times10^{-2})(2.52) =4.03\times10^{-2}\,\mathrm{s}. \] The current falls to \(0.500\,\mathrm{A}\) in \(40.3\,\mathrm{ms}\).

Section B: Relativity and Quantum Mechanics

Answer 3. Unstable particles between beamline counters

[15 marks]

A beam of unstable particles travels in the \(+x\) direction at speed \(0.960c\) between two counters. Counter A is at the start of a straight evacuated section and counter B is \(18.0\,\mathrm{m}\) downstream. The particles have proper mean lifetime \(\tau_0=32.0\,\mathrm{ns}\). In the rest frame of one particle, a decay product can be emitted along the beam direction with speed \(0.600c\).

a) Calculate the Lorentz factor and the mean lifetime measured in the laboratory.

[3 marks]

For \(v=0.960c\), \[ \gamma=\frac{1}{\sqrt{1-v^2/c^2}} =\frac{1}{\sqrt{1-0.960^2}} =3.57. \] The proper lifetime is measured in the particle rest frame, so the laboratory mean lifetime is time-dilated: \[ \tau_{\mathrm{lab}}=\gamma\tau_0=(3.57)(32.0\,\mathrm{ns}) =114\,\mathrm{ns}. \]

Answer 3. Unstable particles between beamline counters

[15 marks]

b) Find the laboratory flight time between the counters and compare it with the dilated lifetime.

[3 marks]

The flight time from A to B in the laboratory is \[ t=\frac{d}{v}=\frac{18.0}{0.960(3.00\times10^8)} =6.25\times10^{-8}\,\mathrm{s} =62.5\,\mathrm{ns}. \] This is shorter than the lab mean lifetime of \(114\,\mathrm{ns}\), so a substantial fraction of the beam reaches counter B.

Answer 3. Unstable particles between beamline counters

[15 marks]

c) Calculate the fraction of particles that reach counter B and the fraction that decay between the counters.

[4 marks]

For exponential decay, the survival fraction after laboratory time \(t\) is \[ \frac{N}{N_0}=e^{-t/\tau_{\mathrm{lab}}}. \] Here \[ \frac{t}{\tau_{\mathrm{lab}}}=\frac{62.5}{114}=0.547. \] Thus \[ \frac{N}{N_0}=e^{-0.547}=0.579. \] The fraction that decays before reaching B is \[ 1-0.579=0.421. \] So about \(57.9\%\) reach B and \(42.1\%\) decay between A and B.

Answer 3. Unstable particles between beamline counters

[15 marks]

d) Use relativistic velocity addition to find the laboratory velocities of products emitted forward and backward along the beam in the particle rest frame. State the frames used. The velocity-addition equation is \[ u_x=\frac{u_x'+v}{1+u_x'v/c^2}, \] where \(S'\) is the particle rest frame and \(S\) is the laboratory frame.

[4 marks]

Let \(S'\) be the unstable particle's rest frame and \(S\) be the laboratory. The frame \(S'\) moves at \(v=0.960c\) relative to \(S\). For a decay product emitted forward with \(u_x'=+0.600c\), \[ u_x=\frac{u_x'+v}{1+u_x'v/c^2} =\frac{0.600c+0.960c}{1+(0.600)(0.960)} =0.990c. \] For a product emitted backward with \(u_x'=-0.600c\), \[ u_x=\frac{-0.600c+0.960c}{1-(0.600)(0.960)} =0.849c. \] Even the backward-emitted product moves in the \(+x\) direction in the laboratory because the parent beam is strongly boosted.

Answer 3. Unstable particles between beamline counters

[15 marks]

e) Explain why the counter spacing would be badly underestimated if the proper lifetime were used directly in the laboratory.

[1 mark]

Without time dilation, the mean distance would be \[ d_0=v\tau_0=(0.960)(3.00\times10^8)(32.0\times10^{-9}) =9.22\,\mathrm{m}. \] With time dilation, the mean laboratory distance is \[ d_{\mathrm{lab}}=v\tau_{\mathrm{lab}}=(0.960)(3.00\times10^8)(114\times10^{-9}) =32.9\,\mathrm{m}. \] The detector spacing \(18.0\,\mathrm{m}\) is reasonable only when the dilated lifetime is used.

Answer 4. Electron states on a quantum ring

[15 marks]

An electron is confined to a very thin conducting quantum ring of radius \(R=0.850\,\mathrm{nm}\). Its position around the ring is described by the angle \(\phi\), and the potential energy is constant around the ring. The spatial wavefunction must be single-valued after one complete circuit.

a) State the boundary condition for the ring and explain its physical meaning.

[3 marks]

For motion on a ring, the coordinate \(\phi\) and \(\phi+2\pi\) represent the same physical point. Therefore the wavefunction must satisfy the periodic boundary condition \[ \psi(\phi+2\pi)=\psi(\phi). \] This replaces the hard-wall condition used in a one-dimensional box. The wavefunction does not need to vanish anywhere on an ideal ring; it must be continuous and single-valued around the loop.

Answer 4. Electron states on a quantum ring

[15 marks]

b) Use the periodic boundary condition to find the allowed values of \(m\) and normalize the states. The trial angular states are \[ \psi_m(\phi)=Ae^{im\phi}. \]

[4 marks]

Take trial states \[ \psi_m(\phi)=Ae^{im\phi}. \] The boundary condition gives \[ Ae^{im(\phi+2\pi)}=Ae^{im\phi}, \] so \[ e^{i2\pi m}=1. \] This requires \[ m=0,\pm1,\pm2,\ldots \] The integer \(m\) is the angular-momentum quantum number. Normalizing over \(0\le\phi<2\pi\), \[ 1=\int_0^{2\pi}|A|^2\,d\phi=2\pi|A|^2, \] so a normalized angular state is \[ \psi_m(\phi)=\frac{1}{\sqrt{2\pi}}e^{im\phi}. \]

Answer 4. Electron states on a quantum ring

[15 marks]

c) Find the angular momentum eigenvalue and derive the energy levels. The angular-momentum operator is \[ \hat L_z=-i\hbar\frac{d}{d\phi} \] and the rotational kinetic energy is \[ E=\frac{L_z^2}{2I}, \] where \(I=m_eR^2\).

[3 marks]

The angular momentum operator about the ring axis is \[ \hat L_z=-i\hbar\frac{d}{d\phi}. \] Acting on the state gives \[ \hat L_z\psi_m=-i\hbar(im)\psi_m=m\hbar\psi_m. \] Thus \[ L_z=m\hbar. \] The kinetic energy for rotation is \[ E_m=\frac{L_z^2}{2I}=\frac{m^2\hbar^2}{2I}, \] where the electron's moment of inertia on the ring is \[ I=m_eR^2. \] Therefore \[ E_m=\frac{m^2\hbar^2}{2m_eR^2}. \]

Answer 4. Electron states on a quantum ring

[15 marks]

d) Discuss the degeneracy of the \(+m\) and \(-m\) states and identify any exception.

[2 marks]

The energy depends on \(m^2\), so \[ E_{+m}=E_{-m}. \] The states \(+m\) and \(-m\) are distinct because they have opposite angular momenta, corresponding to opposite circulation directions around the ring. For \(|m|\ge1\), the levels are therefore twofold degenerate. The \(m=0\) state is not paired with a different opposite state, so it is non-degenerate in this simple model.

Answer 4. Electron states on a quantum ring

[15 marks]

e) Calculate \(E_1\) and the transition energy for \(m=2\to m=1\), in joules and electronvolts.

[3 marks]

Using \(\hbar=h/(2\pi)=1.06\times10^{-34}\,\mathrm{J\,s}\), the energy scale is \[ E_1=\frac{\hbar^2}{2m_eR^2}. \] Substitute \(R=0.850\times10^{-9}\,\mathrm{m}\): \[ E_1=\frac{(1.06\times10^{-34})^2}{2(9.11\times10^{-31})(0.850\times10^{-9})^2} =8.53\times10^{-21}\,\mathrm{J}. \] In electronvolts, \[ E_1=\frac{8.53\times10^{-21}}{1.60\times10^{-19}} =5.33\times10^{-2}\,\mathrm{eV}. \] For a transition from \(m=2\) to \(m=1\), \[ \Delta E=E_2-E_1=(4-1)E_1=3E_1. \] Thus \[ \Delta E=2.56\times10^{-20}\,\mathrm{J}=0.160\,\mathrm{eV}. \]