Question 2
2
A sphere of mass \(1.6 \times 10^{-10} \mathrm{~kg}\) has a charge of +0.27 nC . The sphere is in a uniform electric field that acts vertically upwards, as shown in the side view in Fig. 2.1. SIDE VIEW The force exerted on the sphere by the electric field causes the sphere to remain at a fixed vertical height in a horizontal plane. There is a uniform magnetic field in the region of the electric field. The sphere moves at a speed of \(0.78 \mathrm{~m} \mathrm{~s}^{-1}\) in the horizontal plane. The magnetic field causes the sphere to move in a circular path of radius 3.4 m , as shown in the view from above in Fig. 2.2.
structured6 marks
Question 2(a)
2(a)
3 marks
Question 2(a)(i)
2(a)(i)
Determine the direction of the uniform magnetic field.
Mediumstructured1 marks
Answer
(vertically) downwards B1
Question 2(a)(ii)
2(a)(ii)
Explain why the motion of the sphere in the horizontal plane is circular.
Mediumstructured2 marks
Answer
magnetic force (on sphere) is perpendicular to its velocity B1 magnetic force perpendicular to velocity is the centripetal force or magnetic force perpendicular to velocity causes centripetal acceleration or acceleration perpendicular to velocity is centripetal (acceleration) or magnetic force does not change the speed of the sphere or magnetic force has constant magnitude B1
Question 2(c)
2(c)
By considering the magnetic force on the sphere, show that the flux density of the uniform magnetic field is 0.14 T .
Mediumstructured3 marks
Answer
centripetal force = magnetic force or \(B q v=m v^{2} / r\) B1 B=m v / q r C1 \(=\left(1.6 \times 10^{-10} \times 0.78\right) /\left(0.27 \times 10^{-9} \times 3.4\right)=0.14 \mathrm{~T}\) A1
Question 4
4
A proton of mass m and charge +q is travelling through a vacuum in a straight line with speed v. The magnetic field is normal to the direction of motion of the proton.
structured6 marks
Question 4(a)
4(a)
Explain why the path of the proton in the magnetic field is an arc of a circle.
Mediumstructured2 marks
Answer
M1 A1 [2]
Question 4(b)
4(b)
The angular speed of the proton in the magnetic field is \(\omega\). Derive an expression for \(\omega\) in terms of B, q and m.
Mediumstructured4 marks
Answer
magnetic force =B q v centripetal force \(=m r \omega^{2}\) or \(m v^{2} / r\) \(v=r \omega\) \(B q v=B q r \omega=m r \omega^{2}\) \(\omega=B q / m\) B1 B1 B1 A1 [4]
Question 5
5
8 marks
Question 5(b)
5(b)
A charged particle of mass m and charge +q is travelling with velocity v in a vacuum. It enters a region of uniform magnetic field of flux density B, as shown in Fig. 5.1. The magnetic field is normal to the direction of motion of the particle. The path of the particle in the field is the arc of a circle of radius r.
structured4 marks
Question 5(b)(i)
5(b)(i)
Explain why the path of the particle in the field is the arc of a circle.
Mediumstructured2 marks
Answer
force on particle is (always) normal to velocity / direction of travel B1 speed of particle is constant B1
Question 5(b)(ii)
5(b)(ii)
Show that the radius r is given by the expression
Mediumstructured2 marks
Answer
magnetic force provides the centripetal force B1 \(m v^{2} / r=B q v \quad\) M1 r=m v / B q A0
Question 5(c)
5(c)
A thin metal foil is placed in the magnetic field in (b). A second charged particle enters the region of the magnetic field. It loses kinetic energy as it passes through the foil. The particle follows the path shown in Fig. 5.2.
structured4 marks
Question 5(c)(i)
5(c)(i)
On Fig. 5.2, mark with an arrow the direction of travel of the particle.
Easystructured1 marks
Answer
direction from 'bottom to top' of diagram B1
Question 5(c)(ii)
5(c)(ii)
The path of the particle has different radii on each side of the foil. The radii are 7.4 cm and 5.7 cm . Determine the ratio for the particle as it passes through the foil.
Hardstructured3 marks
Answer
radius proportional to momentum C1 ratio = 5.7 / 7.4 =0.77 A1 (answer must be consistent with direction given in (c)(i))
Question 6
6
6 marks
Question 6(a)
6(a)
Explain the use of a uniform electric field and a uniform magnetic field for the selection of the velocity of a charged particle. You may draw a diagram if you wish.
Hardstructured3 marks
Answer
electric and magnetic fields normal to each other either charged particle enters region normal to both fields or correct B direction w.r.t. E for zero deflection B1 for no deflection, v=E / B (no credit if magnetic field region clearly not overlapping with electric field region)
Question 6(b)
6(b)
Ions, all of the same isotope, are travelling in a vacuum with a speed of \(9.6 \times 10^{4} \mathrm{~ms}^{-1}\). The ions are incident normally on a uniform magnetic field of flux density 640 mT . The ions follow semicircular paths A and B before reaching a detector, as shown in Fig. 6.1. Fig. 6.1 Data for the diameters of the paths are shown in Fig. 6.2. \begin{tabular}{|c|c|} \hline path & diameter \(/ \mathrm{cm}\) \\ \hline A & 6.2 \\ B & 12.4 \\ \hline \end{tabular} The ions in path B each have charge \(+1.6 \times 10^{-19} \mathrm{C}\).
structured1 marks
Question 6(b)(ii)
6(b)(ii)
Suggest and explain quantitatively a reason for the difference in radii of the paths A and B of the ions.
Hardstructured3 marks
Answer
\(q / m \propto 1 / r\) or m constant and \(q \propto 1 / r\) B1 q / m for A is twice that for \(\mathrm{B} \quad \mathrm{B} 1\) ions in path A have (same mass but) twice the charge (of ions in path B ) \(\quad B1 \quad[3]\)
Question 7
7
A solenoid is connected in series with a battery and a switch. A Hall probe is placed close to one end of the solenoid, as illustrated in Fig. 7.1. The current in the solenoid is switched on. The Hall probe is adjusted in position to give the maximum reading. The current is then switched off.
structured3 marks
Question 7(a)
7(a)
The current in the solenoid is now switched on again. Several seconds later, it is switched off. The Hall probe is not moved. On the axes of Fig. 7.2, sketch a graph to show the variation with time t of the Hall voltage \(V_{\mathrm{H}}\).
Mediumstructured3 marks
Answer
graph: \(V_{\mathrm{H}}\) increases from zero when current switched on B1 \(V_{\mathrm{H}}\) then non-zero constant B1 \(V_{\mathrm{H}}\) returns to zero when current switched off B1