Question 1
(a)(i) Define gravitational field. [1 mark]
………………………………………………………………………………………………………………………….
(ii) Define electric field. [1 mark]
………………………………………………………………………………………………………………………….
(iii) State one similarity and one difference between the gravitational potential due to a point mass and the electric potential due to a point charge.
similarity: ………………………………………………………………………………………………………………..
difference: ………………………………………………………………………………………………………………. [2 marks]
(b) An isolated uniform conducting sphere has mass \(M\) and charge \(Q\).
The gravitational field strength at the surface of the sphere is \(g\).
The electric field strength at the surface of the sphere is \(E\).
(i) Show that
\( \frac{M}{Q}=\alpha\frac{g}{E} \)
where \(\alpha\) is a constant. [3 marks]
(ii) Show that the numerical value of \(\alpha\) is \(1.35\times10^{20}\,\mathrm{kg^2\,C^{-2}}\). [1 mark]
________________________________
(c) Assume that the Earth is a uniform conducting sphere of mass \(5.98\times10^{24}\,\mathrm{kg}\).
The surface of the Earth carries a charge of \(-4.80\times10^5\,\mathrm{C}\) that is evenly distributed.
(i) Use the information in (b) to determine the electric field strength at the surface of the Earth. Give a unit with your answer.
electric field strength = ……………………………. unit …………… [2 marks]
(ii) State how the direction of the electric field at the surface of the Earth compares with the direction of the gravitational field.
…………………………………………………………………………………………………………………………. [1 mark]
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 18.1: Electric fields and field lines — part (a)(ii)
• 13.4: Gravitational potential — part (a)(iii)
• 18.5: Electric potential — part (a)(iii)
• 13.3: Gravitational field of a point mass — parts (b)(i), (b)(ii) and (c)(i)
• 18.4: Electric field of a point charge — parts (b)(i), (b)(ii) and (c)(i)
▶️ Answer/Explanation
(a)(i) Gravitational field [1 mark]
A gravitational field is a region in which a mass experiences a gravitational force.
Gravitational field strength is defined as the force per unit mass:
\( g=\frac{F}{m} \)
Answer: \( \boxed{\text{force per unit mass}} \)
(a)(ii) Electric field [1 mark]
An electric field is a region in which a charge experiences an electric force.
Electric field strength is defined as the force per unit positive charge:
\( E=\frac{F}{Q} \)
Answer: \( \boxed{\text{force per unit positive charge}} \)
(a)(iii) Similarity and difference [2 marks]
Similarity: Both gravitational potential and electric potential are inversely proportional to distance from the point source.
Difference: Gravitational potential is always negative, whereas electric potential can be positive or negative.
Answer: \( \boxed{\text{any one valid similarity and one valid difference}} \)
(b)(i) Derivation of the relationship [3 marks]
For the gravitational field at the surface of a spherical mass,
\( g=\frac{GM}{r^2} \)
For the electric field at the surface of a charged sphere,
\( E=\frac{Q}{4\pi\varepsilon_0r^2} \)
Rearranging the gravitational field equation gives
\( M=\frac{gr^2}{G} \)
Rearranging the electric field equation gives
\( Q=4\pi\varepsilon_0Er^2 \)
Therefore,
\( \frac{M}{Q}=\frac{gr^2/G}{4\pi\varepsilon_0Er^2} \)
\( \frac{M}{Q}=\frac{1}{4\pi G\varepsilon_0}\frac{g}{E} \)
Comparing with \( \frac{M}{Q}=\alpha\frac{g}{E} \),
\( \alpha=\frac{1}{4\pi G\varepsilon_0} \)
Answer: \( \boxed{\alpha=\frac{1}{4\pi G\varepsilon_0}} \)
(b)(ii) Numerical value of \(\alpha\) [1 mark]
Using \(G=6.67\times10^{-11}\,\mathrm{N\,m^2\,kg^{-2}}\) and \(\varepsilon_0=8.85\times10^{-12}\,\mathrm{F\,m^{-1}}\),
\( \alpha=\frac{1}{4\pi(6.67\times10^{-11})(8.85\times10^{-12})} \)
\( \alpha=1.35\times10^{20}\,\mathrm{kg^2\,C^{-2}} \)
Answer: \( \boxed{1.35\times10^{20}\,\mathrm{kg^2\,C^{-2}}} \)
(c)(i) Electric field strength at the Earth’s surface [2 marks]
From part (b),
\( \frac{M}{Q}=\alpha\frac{g}{E} \)
Rearranging for \(E\),
\( E=\frac{\alpha gQ}{M} \)
Using \( \alpha=1.35\times10^{20}\,\mathrm{kg^2\,C^{-2}} \), \(g=9.81\,\mathrm{N\,kg^{-1}}\), \(Q=4.80\times10^5\,\mathrm{C}\) and \(M=5.98\times10^{24}\,\mathrm{kg}\):
\( E=\frac{(1.35\times10^{20})(9.81)(4.80\times10^5)}{5.98\times10^{24}} \)
\( E=1.06\times10^2\,\mathrm{N\,C^{-1}} \)
Answer: \( \boxed{1.06\times10^2\,\mathrm{N\,C^{-1}}} \)
(c)(ii) Direction [1 mark]
The Earth has a negative charge, so the electric field is directed towards the Earth.
The gravitational field is also directed towards the Earth.
Answer: \( \boxed{\text{same direction}} \)
Question 2
A steel sphere of mass \(0.29\,\mathrm{kg}\) is suspended in equilibrium from a vertical spring. The centre of the sphere is \(8.5\,\mathrm{cm}\) from the top of the spring, as shown in Fig. 2.1.
The sphere is now set in motion so that it is moving in a horizontal circle at constant speed, as shown in Fig. 2.2.
The distance from the centre of the sphere to the top of the spring is now \(10.8\,\mathrm{cm}\).
(a) Explain, with reference to the forces acting on the sphere, why the length of the spring in Fig. 2.2 is greater than in Fig. 2.1. [3 marks]
………………………………………………………………………………………………………………………….
(b) The angle between the linear axis of the spring and the vertical is \(27^\circ\).
(i) Show that the radius \(r\) of the circle is \(4.9\,\mathrm{cm}\). [1 mark]
________________________________
(ii) Show that the tension in the spring is \(3.2\,\mathrm{N}\). [2 marks]
________________________________
(iii) The spring obeys Hooke’s law.
Calculate the spring constant, in \(\mathrm{N\,cm^{-1}}\), of the spring.
spring constant = ……………………………………… \( \mathrm{N\,cm^{-1}} \) [2 marks]
(c)
(i) Use the information in (b) to determine the centripetal acceleration of the sphere.
centripetal acceleration = ………………………………………… \( \mathrm{m\,s^{-2}} \) [2 marks]
(ii) Calculate the period of the circular motion of the sphere.
period = ……………………………………………… \( \mathrm{s} \) [2 marks]
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 12.2: Centripetal acceleration — parts (a), (b)(i), (b)(ii) and (c)(i)
• 6.1: Stress and strain — part (b)(iii)
• 12.1: Kinematics of uniform circular motion — part (c)(ii)
▶️ Answer/Explanation
(a) [3 marks]
When the sphere is stationary, the tension in the spring balances the weight:
\( T=mg \)
When the sphere moves in a horizontal circle, the tension in the spring has both vertical and horizontal components.
The vertical component of tension balances the weight, while the horizontal component provides the centripetal force needed for circular motion.
Therefore, the tension is greater than the weight:
\(T\cos27^\circ=mg\)
so \(T>mg\).
The greater tension produces a greater extension of the spring, so the spring is longer in Fig. 2.2.
Answer: \( \boxed{\text{horizontal component provides centripetal force, so tension is greater and the spring extends further}} \)
(b)(i) Radius of the circle [1 mark]
From the geometry of Fig. 2.2,
\( r=10.8\sin27^\circ \)
\( r=4.9\,\mathrm{cm} \)
Answer: \( \boxed{4.9\,\mathrm{cm}} \)
(b)(ii) Tension in the spring [2 marks]
The vertical component of the tension balances the weight:
\(T\cos27^\circ=mg\)
Therefore,
\(T=\frac{mg}{\cos27^\circ}\)
\(T=\frac{(0.29)(9.81)}{\cos27^\circ}\)
\(T=3.2\,\mathrm{N}\)
Answer: \( \boxed{3.2\,\mathrm{N}} \)
(b)(iii) Spring constant [2 marks]
In the initial equilibrium position,
\(T_1=mg=(0.29)(9.81)=2.84\,\mathrm{N}\)
The change in extension is
\(\Delta x=10.8-8.5=2.3\,\mathrm{cm}\)
Using Hooke’s law,
\(k=\frac{\Delta T}{\Delta x}\)
\(k=\frac{3.2-2.84}{2.3}\)
\(k=0.15\,\mathrm{N\,cm^{-1}}\)
Answer: \( \boxed{0.15\,\mathrm{N\,cm^{-1}}} \)
(c)(i) Centripetal acceleration [2 marks]
The horizontal component of the tension provides the centripetal force:
\(ma=T\sin27^\circ\)
Therefore,
\(a=\frac{T\sin27^\circ}{m}\)
\(a=\frac{(3.2)\sin27^\circ}{0.29}\)
\(a=5.0\,\mathrm{m\,s^{-2}}\)
Answer: \( \boxed{5.0\,\mathrm{m\,s^{-2}}} \)
(c)(ii) Period of circular motion [2 marks]
For circular motion,
\(a=r\omega^2\)
and
\(\omega=\frac{2\pi}{T}\)
Hence,
\(a=r\left(\frac{2\pi}{T}\right)^2\)
Rearranging,
\(T=2\pi\sqrt{\frac{r}{a}}\)
Using \(r=4.9\,\mathrm{cm}=0.049\,\mathrm{m}\) and \(a=5.0\,\mathrm{m\,s^{-2}}\):
\(T=2\pi\sqrt{\frac{0.049}{5.0}}\)
\(T=0.62\,\mathrm{s}\)
Answer: \( \boxed{0.62\,\mathrm{s}} \)
Question 3
(a) State the reason why two objects that are at the same temperature are described as being in thermal equilibrium. [1 mark]
………………………………………………………………………………………………………………………….
(b) Fig. 3.1 shows the variations with temperature of the densities of mercury and of water between \(0^\circ\mathrm{C}\) and \(100^\circ\mathrm{C}\).
Temperature may be measured using the variation with temperature of the density of a liquid.
Suggest why, for measuring temperature over this temperature range:
(i) mercury is a suitable liquid.
…………………………………………………………………………………………………………………………. [1 mark]
(ii) water is not a suitable liquid.
…………………………………………………………………………………………………………………………. [2 marks]
(c) A beaker contains a liquid of mass \(120\,\mathrm{g}\). The liquid is supplied with thermal energy at a rate of \(810\,\mathrm{W}\). The beaker has a mass of \(42\,\mathrm{g}\) and a specific heat capacity of \(0.84\,\mathrm{J\,g^{-1}\,K^{-1}}\). The beaker and the liquid are in thermal equilibrium with each other at all times and are insulated from the surroundings.
Fig. 3.2 shows the variation with time \(t\) of the temperature of the liquid.
(i) State the boiling temperature, in \(^\circ\mathrm{C}\), of the liquid.
temperature = ……………………………………………. \(^\circ\mathrm{C}\) [1 mark]
(ii) Determine the specific heat capacity, in \(\mathrm{J\,g^{-1}\,K^{-1}}\), of the liquid.
specific heat capacity = ……………………………………. \( \mathrm{J\,g^{-1}\,K^{-1}} \) [4 marks]
(d) The experiment in (c) is repeated using water instead of the liquid in (c). The mass of liquid used, the power supplied, and the initial temperature are all unchanged. The specific heat capacity of water is approximately twice that of the liquid in (c). The boiling temperature of water is \(100^\circ\mathrm{C}\).
On Fig. 3.2, sketch the variation with time \(t\) of the temperature of the water between \(t=0\) and \(t=60\,\mathrm{s}\). Numerical calculations are not required. [2 marks]
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 14.2: Temperature scales — part (b)
• 14.3: Specific heat capacity and specific latent heat — parts (c) and (d)
▶️ Answer/Explanation
(a) Thermal equilibrium [1 mark]
When two objects are at the same temperature, there is no net thermal energy transfer between them.
Answer: \( \boxed{\text{no net thermal energy is transferred between them}} \)
(b)(i) Mercury as a suitable liquid [1 mark]
The density of mercury varies approximately linearly with temperature over the range \(0^\circ\mathrm{C}\) to \(100^\circ\mathrm{C}\).
Answer: \( \boxed{\text{density varies linearly with temperature}} \)
(b)(ii) Water as an unsuitable liquid [2 marks]
The variation of density of water with temperature is not linear.
Also, there is a region where the density does not change simply with temperature, and different temperatures can correspond to the same density.
Answer: \( \boxed{\text{non-linear variation and non-unique density-temperature relationship}} \)
(c)(i) Boiling temperature [1 mark]
The temperature remains constant at the boiling point. From Fig. 3.2, the boiling temperature is \(80^\circ\mathrm{C}\).
Answer: \( \boxed{80^\circ\mathrm{C}} \)
(c)(ii) Specific heat capacity [4 marks]
From the graph, the temperature increases from \(25^\circ\mathrm{C}\) to \(80^\circ\mathrm{C}\) in approximately \(21\,\mathrm{s}\).
The thermal energy supplied is
\(Q=Pt\)
\(Q=(810)(21)=1.70\times10^4\,\mathrm{J}\)
Some of this energy heats the beaker.
\(Q_{\mathrm{beaker}}=mc\Delta T\)
\(Q_{\mathrm{beaker}}=(42)(0.84)(80-25)\)
\(Q_{\mathrm{beaker}}\approx1.94\times10^3\,\mathrm{J}\)
Therefore, the energy absorbed by the liquid is
\(Q_{\mathrm{liquid}}=(810\times21)-(42\times0.84\times55)\)
Using \(Q=mc\Delta T\),
\(c=\frac{Q_{\mathrm{liquid}}}{m\Delta T}\)
\(c=\frac{(810\times21)-(42\times0.84\times55)}{120\times55}\)
\(c\approx2.3\,\mathrm{J\,g^{-1}\,K^{-1}}\)
Answer: \( \boxed{2.3\,\mathrm{J\,g^{-1}\,K^{-1}}} \)
(d) Temperature-time graph for water [2 marks]
The specific heat capacity of water is approximately twice that of the original liquid.
For the same mass and power input, \(P=mc\frac{\Delta T}{\Delta t}\), so doubling \(c\) makes the temperature-time gradient approximately half as large.
The graph should therefore start at \(25^\circ\mathrm{C}\) and rise as a straight line with approximately half the original gradient until it reaches \(100^\circ\mathrm{C}\).
It should then become horizontal at \(100^\circ\mathrm{C}\) because the water is boiling.
Answer: \( \boxed{\text{straight line from }25^\circ\mathrm{C}\text{ with half the original gradient, then horizontal at }100^\circ\mathrm{C}} \)
Question 4
(a) State two of the basic assumptions of the kinetic theory of gases. [2 marks]
1. ………………………………………………………………………………………………………………………….
2. ………………………………………………………………………………………………………………………….
(b) An ideal gas has amount of substance \(n\).
The gas is initially in state X, with pressure \(2p\) and volume \(V\).
The gas is cooled at constant volume to state Y, with pressure \(p\).
The gas is then heated at constant pressure to state Z, with volume \(2V\).
Finally, the gas returns at constant temperature to state X.
(i) Determine an expression for the temperature \(T\) of the gas in state X, in terms of \(n\), \(p\) and \(V\).
Identify any other symbols that you use. [2 marks]
(ii) On Fig. 4.1, sketch the variation with volume of pressure for the gas as the gas undergoes the three changes. The state X is labelled. Label states Y and Z.[3 marks]
(iii) During the change of state from Y to Z, the increase in internal energy of the gas is \(U\).
During the change of state from Z to X, the work done on the gas is \(W\).
Complete Table 4.1 to indicate, for each of the three changes of state, the increase in internal energy of the gas, the thermal energy transferred to the gas and the work done on the gas, in terms of \(p\), \(V\), \(U\) and \(W\).[5 marks]
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 15.1: Equation of state — parts (b)(i) and (b)(ii)
• 16.1: Internal energy — part (b)(iii)
• 16.2: First law of thermodynamics — part (b)(iii)
▶️ Answer/Explanation
(a) Kinetic theory assumptions [2 marks]
Any two of the following:
• Gas particles are in continuous random motion.
• The volume of the particles is negligible compared with the volume occupied by the gas.
• There are negligible forces between particles except during collisions.
• Collisions between particles and with the container walls are perfectly elastic.
• The time taken for a collision is negligible compared with the time between collisions.
Answer: Any two valid assumptions.
(b)(i) Temperature of state X [2 marks]
For an ideal gas,
\(pV=nRT\)
At state X, the pressure is \(2p\) and the volume is \(V\), so
\(2pV=nRT\)
Therefore,
\(T=\frac{2pV}{nR}\)
where \(R\) is the molar gas constant.
Answer: \( \boxed{T=\frac{2pV}{nR}} \), where \(R\) is the molar gas constant.
(b)(ii) Pressure-volume graph [3 marks]
For \(X\rightarrow Y\), the volume remains constant at \(V\), while the pressure decreases from \(2p\) to \(p\). This is a vertical line.
For \(Y\rightarrow Z\), the pressure remains constant at \(p\), while the volume increases from \(V\) to \(2V\). This is a horizontal line.
For \(Z\rightarrow X\), the temperature remains constant, so \(pV=\text{constant}\). Hence the graph is an isothermal curve from \((2V,p)\) to \((V,2p)\).
Required graph: vertical \(XY\), horizontal \(YZ\), followed by a curved isothermal path \(ZX\).
(b)(iii) Energy changes [5 marks]
The first law is
\(\Delta U=q+W\)
where \(q\) is the thermal energy transferred to the gas and \(W\) is the work done on the gas.
For \(X\rightarrow Y\):
The volume is constant, so the work done on the gas is \(0\).
Hence \(W=0\) and \(\Delta U=q\).
Since the gas subsequently returns to its initial state, the total change in internal energy around the complete cycle is zero.
For \(Y\rightarrow Z\):
The increase in internal energy is \(+U\).
The gas expands from \(V\) to \(2V\) at constant pressure \(p\), so the work done on the gas is
\(W=-p\Delta V=-pV\)
Therefore, from \(\Delta U=q+W\),
\(U=q-pV\)
so
\(q=U+pV\)
For \(Z\rightarrow X\):
The temperature is constant, so the internal energy of an ideal gas is unchanged:
\(\Delta U=0\)
The work done on the gas is given as \(+W\).
Therefore,
\(0=q+W\)
so
\(q=-W\)
Completed values:
\(X\rightarrow Y:\quad \Delta U=-U,\quad q=-U,\quad W=0\)
\(Y\rightarrow Z:\quad \Delta U=+U,\quad q=U+pV,\quad W=-pV\)
\(Z\rightarrow X:\quad \Delta U=0,\quad q=-W,\quad W=+W\)
Answer: \( \boxed{\Delta U=q+W} \) must be satisfied for each individual process.
Final table:
| Change | Increase in internal energy | Thermal energy transferred to gas | Work done on gas |
|---|---|---|---|
| \(X\rightarrow Y\) | \(-U\) | \(-U\) | \(0\) |
| \(Y\rightarrow Z\) | \(+U\) | \(U+pV\) | \(-pV\) |
| \(Z\rightarrow X\) | \(0\) | \(-W\) | \(+W\) |
Question 5
Part of an electric circuit is shown in Fig. 5.1.
The circuit is used to produce half-wave rectification of an alternating voltage of potential difference (p.d.) \(V_{\mathrm{IN}}\).
The output p.d. across the \(14\,\mathrm{k\Omega}\) resistor is \(V_{\mathrm{OUT}}\).
(a)(i) A component is missing from the circuit of Fig. 5.1.
Complete the circuit diagram in Fig. 5.1 by adding the circuit symbol for the missing component, correctly connected. [1 mark]
(ii) A capacitor \(C\) is shown in the circuit of Fig. 5.1.
State the effect on \(V_{\mathrm{OUT}}\) of including the capacitor in the circuit. [1 mark]
(b) Fig. 5.2 shows the variation with time \(t\) of \(V_{\mathrm{IN}}\).
Fig. 5.3 shows the variation with \(t\) of \(V_{\mathrm{OUT}}\).
(i) Determine the frequency of \(V_{\mathrm{IN}}\).
frequency = ……………………………………………. \( \mathrm{Hz} \) [1 mark]
(ii) Show that the time constant \(\tau\) for the discharge of the capacitor through the resistor is \(0.038\,\mathrm{s}\). [2 marks]
________________________________
(iii) Calculate the capacitance of \(C\). Give a unit with your answer.
capacitance = ……………………………. unit …………… [2 marks]
(c) The circuit of Fig. 5.1 is modified so that it produces full-wave rectification of an input voltage.
Suggest, with a reason, how \(V_{\mathrm{OUT}}\) now varies with time when \(V_{\mathrm{IN}}\) is as shown in Fig. 5.2. [2 marks]
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 21.1: Characteristics of alternating currents — part (b)(i)
• 19.3: Discharging a capacitor — parts (b)(ii) and (b)(iii)
▶️ Answer/Explanation
(a)(i) Missing component [1 mark]
The missing component is a diode, connected in series in the upper branch of the circuit.
Answer: \( \boxed{\text{diode}} \)
(a)(ii) Effect of the capacitor [1 mark]
The capacitor charges during the conducting part of the cycle and discharges through the \(14\,\mathrm{k\Omega}\) resistor when the diode is not conducting.
This reduces the variation in \(V_{\mathrm{OUT}}\), producing a smoother output.
Answer: \( \boxed{V_{\mathrm{OUT}}\text{ is smoothed}} \)
(b)(i) Frequency of \(V_{\mathrm{IN}}\) [1 mark]
From Fig. 5.2, one complete cycle takes \(0.040\,\mathrm{s}\).
Using
\(f=\frac{1}{T}\)
\(f=\frac{1}{0.040}\)
\(f=25\,\mathrm{Hz}\)
Answer: \( \boxed{25\,\mathrm{Hz}} \)
(b)(ii) Time constant [2 marks]
For the discharge of a capacitor,
\(V=V_0e^{-t/\tau}\)
where \(\tau=RC\).
From Fig. 5.3, \(V_0=5.50\,\mathrm{V}\), \(V=3.25\,\mathrm{V}\) at \(t=0.020\,\mathrm{s}\).
Therefore,
\(3.25=5.50e^{-0.020/\tau}\)
Solving gives
\(\tau\approx0.038\,\mathrm{s}\)
Answer: \( \boxed{0.038\,\mathrm{s}} \)
(b)(iii) Capacitance [2 marks]
The time constant is
\(\tau=RC\)
Hence
\(C=\frac{\tau}{R}\)
\(C=\frac{0.038}{14000}\)
\(C=2.7\times10^{-6}\,\mathrm{F}\)
Answer: \( \boxed{2.7\times10^{-6}\,\mathrm{F}} \)
(c) Full-wave rectification [2 marks]
With full-wave rectification, both the positive and negative half-cycles of \(V_{\mathrm{IN}}\) produce an output of the same polarity.
Therefore, the magnitude of \(V_{\mathrm{IN}}\) is used for both half-cycles, so \(V_{\mathrm{OUT}}\) varies at twice the input frequency.
Answer: \( \boxed{V_{\mathrm{OUT}}\text{ is produced during both half-cycles, giving a frequency of }2f} \)
Question 6
(a) State what is meant by a magnetic field. [2 marks]
………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………
(b) A long, straight wire P carries a current into the page, as shown in Fig. 6.1.
Fig. 6.1
On Fig. 6.1, draw four field lines to represent the magnetic field around wire P due to the current in the wire. [3 marks]
(c) A second long, straight wire Q, carrying a current of \(5.0\,\mathrm{A}\) out of the page, is placed parallel to wire P, as shown in Fig. 6.2.
The flux density of the magnetic field at wire Q due to the current in wire P is \(2.6\,\mathrm{mT}\).
(i) Calculate the magnetic force per unit length exerted on wire Q by wire P. [2 marks]
force per unit length = ……………………………………….. \( \mathrm{N\,m^{-1}} \)
(ii) State the direction of the force exerted on wire Q by wire P. [1 mark]
………………………………………………………………………………………………………………………………
(iii) The flux density of the magnetic field at wire P due to the current in wire Q is \(1.5\,\mathrm{mT}\).
Determine the magnitude of the current in wire P. Explain your reasoning. [2 marks]
current = ……………………………………………… \( \mathrm{A} \)
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 20.4: Magnetic fields due to currents — parts (b) and (c)
• 20.2: Force on a current-carrying conductor — parts (c)(i), (c)(ii) and (c)(iii)
▶️ Answer/Explanation
(a) Magnetic field [2 marks]
A magnetic field is a region where a force acts on a current-carrying conductor.
It may also be a region where a force acts on a moving charge.
A magnetic field can also exert a force on a magnetic material or magnetic pole.
Answer: \( \boxed{\text{a region where a magnetic force acts}} \)
(b) Magnetic field around a straight current-carrying wire [3 marks]
The magnetic field lines around a long, straight wire are concentric circles centred on the wire.
Since the current in wire P is into the page, the arrows on the field lines are in the clockwise direction.
The spacing between the circles increases with distance from the wire.
(c)(i) Magnetic force per unit length [2 marks]
The force on a current-carrying conductor is
\(F=BIL\)
Therefore, the force per unit length is
\(\frac{F}{L}=BI\)
\(B=2.6\times10^{-3}\,\mathrm{T}\) and \(I=5.0\,\mathrm{A}\).
\(\frac{F}{L}=(2.6\times10^{-3})(5.0)\)
\(\frac{F}{L}=0.013\,\mathrm{N\,m^{-1}}\)
Answer: \( \boxed{0.013\,\mathrm{N\,m^{-1}}} \)
(c)(ii) Direction of force [1 mark]
The currents in wires P and Q are in opposite directions, so the wires repel each other.
Answer: \( \boxed{\text{to the right}} \)
(c)(iii) Current in wire P [2 marks]
By Newton’s third law, the force per unit length exerted by wire Q on wire P has the same magnitude as the force per unit length exerted by wire P on wire Q.
Therefore,
\(0.013=(1.5\times10^{-3})I\)
\(I=\frac{0.013}{1.5\times10^{-3}}\)
\(I=8.7\,\mathrm{A}\)
Answer: \( \boxed{8.7\,\mathrm{A}} \)
Question 7
(a) State what is meant by the de Broglie wavelength. [1 mark]
………………………………………………………………………………………………………………………………
(b) Fig. 7.1 shows a glass tube in which electrons are accelerated through a high p.d. to form a beam that is incident on a thin graphite crystal.
After passing through the graphite crystal, the electrons reach the fluorescent screen. The screen glows where the electrons strike it.
Fig. 7.2 shows the fluorescent screen viewed end-on, from the right-hand side of Fig. 7.1.
(i) State the name of the phenomenon demonstrated by the pattern shown in Fig. 7.2. [1 mark]
………………………………………………………………………………………………………………………………
(ii) Explain what can be concluded from the pattern in Fig. 7.2 about the nature of electrons. [2 marks]
………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………
(c) The electrons in (b) are now accelerated through a greater potential difference between the cathode and the anode.
(i) On Fig. 7.3, sketch the pattern that is now seen on the fluorescent screen in Fig. 7.1. [2 marks]
(ii) Explain, with reference to the de Broglie wavelength, the change in the pattern on the fluorescent screen. [3 marks]
………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
▶️ Answer/Explanation
(a) [1 mark]
The de Broglie wavelength is the wavelength associated with a moving particle.
It is given by \( \lambda=\frac{h}{p} \), where \(h\) is the Planck constant and \(p\) is the momentum of the particle.
Answer: \( \boxed{\text{wavelength associated with a moving particle}} \)
(b)(i) [1 mark]
The pattern is produced by electron diffraction.
Answer: \( \boxed{\text{electron diffraction}} \)
(b)(ii) [2 marks]
The beam spreads out after passing through the graphite crystal, indicating diffraction.
The light and dark regions indicate an interference pattern.
Therefore, the electrons exhibit wave-like behaviour.
Answer: The electron beam behaves as a wave.
(c)(i) [2 marks]
The new pattern consists of a central bright spot and concentric rings.
The rings are closer together than in the original pattern.
(c)(ii) [3 marks]
A greater accelerating p.d. gives the electrons greater kinetic energy and therefore greater momentum.
From the de Broglie equation, \( \lambda=\frac{h}{p} \), greater momentum means a smaller de Broglie wavelength.
For the same spacing of the graphite crystal, the smaller wavelength produces a smaller diffraction angle.
Hence the intensity maxima occur closer to the centre and the fringe spacing decreases.
Answer: greater p.d. \( \rightarrow \) greater momentum \( \rightarrow \) smaller \( \lambda \) \( \rightarrow \) smaller diffraction angle \( \rightarrow \) rings closer together.
Question 8
(a) Table 8.1 shows some data relating to the properties of air, gel and body tissue. The data are given to three significant figures.
(i) Show that the specific acoustic impedance of gel is \(1.68\times10^6\,\mathrm{kg\,m^{-2}\,s^{-1}}\). [1 mark]
________________________________
(ii) Complete Table 8.1 by calculating the missing values to three significant figures. Use the space below for any working that you need. [2 marks]
________________________________
(b) Use the information in (a) to calculate the intensity reflection coefficient for:
(i) an air–tissue boundary [2 marks]
intensity reflection coefficient = …………………………………………………
(ii) a gel–tissue boundary [1 mark]
intensity reflection coefficient = …………………………………………………
(c) Use your answers in (b) to explain why gel is applied to the skin during ultrasound scanning. [2 marks]
________________________________
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
▶️ Answer/Explanation
(a)(i) Specific acoustic impedance of gel [1 mark]
Specific acoustic impedance is
\( Z=\rho c \)
For gel, \( \rho=1200\,\mathrm{kg\,m^{-3}} \) and \( c=1400\,\mathrm{m\,s^{-1}} \).
\( Z=(1200)(1400) \)
\( Z=1.68\times10^6\,\mathrm{kg\,m^{-2}\,s^{-1}} \)
Answer: \( \boxed{1.68\times10^6\,\mathrm{kg\,m^{-2}\,s^{-1}}} \)
(a)(ii) Missing values [2 marks]
For air, \( Z=\rho c \).
\( \rho=\frac{Z}{c}=\frac{440}{340}=1.29\,\mathrm{kg\,m^{-3}} \)
For tissue, \( Z=\rho c \).
\( c=\frac{Z}{\rho}=\frac{1.68\times10^6}{1090} \)
\( c=1.54\times10^3\,\mathrm{m\,s^{-1}} \)
Answers: \( \boxed{1.29\,\mathrm{kg\,m^{-3}}} \), \( \boxed{1.54\times10^3\,\mathrm{m\,s^{-1}}} \)
(b)(i) Air–tissue boundary [2 marks]
The intensity reflection coefficient is
\( \frac{I_{\mathrm{R}}}{I_0}=\left(\frac{Z_1-Z_2}{Z_1+Z_2}\right)^2 \)
Using \(Z_{\mathrm{air}}=440\,\mathrm{kg\,m^{-2}\,s^{-1}}\) and \(Z_{\mathrm{tissue}}=1.68\times10^6\,\mathrm{kg\,m^{-2}\,s^{-1}}\):
\( \frac{I_{\mathrm{R}}}{I_0}=\left(\frac{1.68\times10^6-440}{1.68\times10^6+440}\right)^2 \)
\( \frac{I_{\mathrm{R}}}{I_0}\approx0.999 \)
Answer: \( \boxed{0.999} \)
(b)(ii) Gel–tissue boundary [1 mark]
Both gel and tissue have the same specific acoustic impedance:
\( Z_{\mathrm{gel}}=Z_{\mathrm{tissue}}=1.68\times10^6\,\mathrm{kg\,m^{-2}\,s^{-1}} \)
Therefore, \( \frac{I_{\mathrm{R}}}{I_0}=\left(\frac{Z_1-Z_2}{Z_1+Z_2}\right)^2=0 \)
Answer: \( \boxed{0} \)
(c) Why gel is applied to the skin [2 marks]
At an air–tissue boundary, the specific acoustic impedances are very different, so almost all of the incident ultrasound is reflected.
The gel has an acoustic impedance closely matched to that of body tissue, giving a very small reflection coefficient. Therefore, much more ultrasound is transmitted into the body.
Answer: The gel removes the air gap between the probe and skin, reducing reflection and allowing more ultrasound to enter the body.
Question 9
Carbon-11 is radioactive and decays by \( \beta^+ \) emission to form boron-11. Carbon-11 has a half-life of 20 minutes. Boron-11 is stable.
(a) Define half-life. [1 mark]
………………………………………………………………………………………………………………………………
(b) A sample contains \(N_0\) nuclei of carbon-11 and no other nuclei at time \(t=0\).
On Fig. 9.1, sketch the variation with \(t\) of the number of nuclei of boron-11 in the sample.
[3 marks]
(c)
(i) Explain, with reference to the random nature of radioactive decay, why the activity of the carbon-11 sample in (b) decreases with time. [2 marks]
………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………
(ii) State, with reasons, whether a radiation detector placed near to the sample of carbon-11 indicates a measured count rate from the sample that is less than, the same as or greater than the activity of the sample. [3 marks]
……………………………………………………………………
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
▶️ Answer/Explanation
(a) Definition of half-life [1 mark]
Half-life is the time taken for the activity of a radioactive sample to decrease to half its initial value.
Answer: \( \boxed{\text{time for the activity of a sample to halve}} \)
(b) Number of boron-11 nuclei [3 marks]
Initially, there are no boron-11 nuclei, so the graph starts at \( (0,0) \).
As carbon-11 decays, boron-11 is produced. The number of boron-11 nuclei therefore increases with time.
The rate of production decreases as fewer undecayed carbon-11 nuclei remain, so the graph has a decreasing gradient.
After one half-life of 20 min, half of the original carbon-11 nuclei have decayed: \( N_{\mathrm{B}}=0.5N_0 \).
After 40 min, three-quarters have decayed: \( N_{\mathrm{B}}=0.75N_0 \).
Therefore, the required graph is an increasing exponential curve starting at \(0\), passing through \( (20,0.5N_0) \) and \( (40,0.75N_0) \), and approaching \(N_0\).
Answer: Increasing exponential curve approaching \(N_0\).
(c)(i) Activity decreases with time [2 marks]
Radioactive decay is random, but every undecayed nucleus has the same probability of decaying in a given time interval.
As time passes, the number of undecayed carbon-11 nuclei decreases. Therefore, fewer nuclei decay in a given time interval, so the activity decreases.
Answer: Fewer undecayed nuclei remain with time, so fewer decays occur per unit time and the activity decreases.
(c)(ii) Count rate compared with activity [3 marks]
The measured count rate is less than the activity.
The radioactive sample emits particles in all directions, but the detector only detects particles travelling towards it.
Some emissions may also be absorbed before reaching the detector, and some may be scattered within the sample.
Therefore, not every decay produces a count in the detector.
Answer: \( \boxed{\text{measured count rate is less than the activity}} \).
Question 10
(a) State Hubble’s law. Identify any symbols that you use. [2 marks]
………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………
(b) A star of luminosity \(3.8\times10^{31}\,\mathrm{W}\) is a distance of \(1.8\times10^{24}\,\mathrm{m}\) from the Earth.
Calculate the radiant flux intensity at the Earth of the radiation emitted by the star. [2 marks]
radiant flux intensity = ………………………………………. \( \mathrm{W\,m^{-2}} \)
(c) The star in (b) is in a distant galaxy. A spectral line in the light from this galaxy is known to have a wavelength of \(486\,\mathrm{nm}\). This spectral line in the light from the galaxy observed on the Earth has a wavelength of \(492\,\mathrm{nm}\).
(i) Explain why the wavelength observed on the Earth is different from the wavelength that the galaxy is known to have emitted. [2 marks]
………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………
(ii) Determine a value for the Hubble constant \(H_0\). [3 marks]
\(H_0\) = …………………………………………… \( \mathrm{s^{-1}} \)
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 25.3: Hubble’s law and the Big Bang theory – parts (a), (c)(i) and (c)(ii)
▶️ Answer/Explanation
(a) Hubble’s law [2 marks]
Hubble’s law states that the speed of recession of a galaxy is directly proportional to its distance from the observer.
\( v=H_0d \)
where \(v\) is the recession speed of the galaxy, \(H_0\) is the Hubble constant and \(d\) is the distance of the galaxy from the observer.
Answer: \( \boxed{v=H_0d} \)
(b) Radiant flux intensity [2 marks]
The radiant flux intensity is given by the inverse-square law:
\( F=\frac{L}{4\pi d^2} \)
\( F=\frac{3.8\times10^{31}}{4\pi(1.8\times10^{24})^2} \)
\( F=9.3\times10^{-19}\,\mathrm{W\,m^{-2}} \)
Answer: \( \boxed{9.3\times10^{-19}\,\mathrm{W\,m^{-2}}} \)
(c)(i) Difference in observed wavelength [2 marks]
The observed wavelength is greater than the emitted wavelength: \(492\,\mathrm{nm}>486\,\mathrm{nm}\).
The galaxy is moving away from the Earth, so the wavelength of the light is increased by the Doppler effect. This is known as redshift.
Answer: The galaxy is receding from the Earth, causing the emitted light to be redshifted.
(c)(ii) Hubble constant [3 marks]
For small redshifts,
\( \frac{\Delta\lambda}{\lambda}=\frac{v}{c} \)
Therefore,
\( v=\frac{(492-486)(3.00\times10^8)}{486} \)
\( v=3.7\times10^6\,\mathrm{m\,s^{-1}} \)
Using Hubble’s law, \( H_0=\frac{v}{d} \)
\( H_0=\frac{3.7\times10^6}{1.8\times10^{24}} \)
\( H_0=2.1\times10^{-18}\,\mathrm{s^{-1}} \)
Answer: \( \boxed{2.1\times10^{-18}\,\mathrm{s^{-1}}} \)