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A-Level CAIE Physics 20 3 Force On A Moving Charge Question Bank

Practice A-Level CAIE Physics 20 3 Force On A Moving Charge questions by syllabus topic with past-paper context, marks, difficulty and question previews on Eduninja.

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Question 2

2

2 marks

Question 2(b)

2(b)

On the axes of Fig. 2.2, sketch the variation with speed of the magnitude of the force on a charged particle moving at right-angles to a uniform magnetic field.

Mediumstructured2 marks

Answer

straight line with positive gradient line starts at origin

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 2

2

2 marks

Question 2(b)

2(b)

Positronium is a system in which an electron and a positron orbit, with the same period, around their common centre of mass, as shown in Fig. 2.1. The radius r of the orbit of both particles is \(1.59 \times 10^{-10} \mathrm{~m}\).

structured2 marks

Question 2(b)(i)

2(b)(i)

Explain how the electric force between the electron and the positron causes the path of the moving particles to be circular.

Hardstructured2 marks

Answer

(electric) force is perpendicular to velocity (of particles) B1 force (perpendicular to velocity) causes centripetal acceleration or force does not change the speed of the particles or force has constant magnitude B1

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 4

4

1 marks

Question 4(c)

4(c)

An electron enters the region between two parallel plates P and Q, that are separated by a distance of 18 mm , as shown in Fig. 4.3. The space between the plates is a vacuum. The potential difference between the plates is 250 V . The electric field may be assumed to be uniform in the region between the plates and zero outside this region.

structured1 marks

Question 4(c)(iii)

4(c)(iii)

Explain why the electron does not follow a circular path.

Hardstructured1 marks

Answer

either the force is not (always) perpendicular to the velocity or the force is always in the same direction B1

Question 5

5

2 marks

Question 5(b)

5(b)

A large horseshoe magnet has a uniform magnetic field between its poles. The magnetic field is zero outside the space between the poles. A small Hall probe is moved at constant speed along a line XY that is midway between, and parallel to, the faces of the poles of the magnet, as shown in Fig. 5.1. An e.m.f. is produced by the Hall probe when it is in the magnetic field. The angle between the plane of the probe and the direction of the magnetic field is not varied. On the axes of Fig. 5.2, sketch a graph to show the variation with time t of the e.m.f. \(V_{\mathrm{H}}\) produced by the Hall probe.

Mediumstructured2 marks

Answer

graph: \(V_{\mathrm{H}\) constant and non zero between the poles and zero outside sharp increase/decrease at ends of magnet} M1 A1 [2]

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

Question 5

5

8 marks

Question 5(b)

5(b)

A charged particle of mass m and with charge q enters a region of uniform magnetic field, perpendicular to the field lines. The magnetic flux density is B. The particle travels in a circle with period T and radius r.

structured8 marks

Question 5(b)(i)

5(b)(i)

By considering the magnetic force acting on the particle, show that

Mediumstructured3 marks

Answer

\(B q v=m v^{2} / r\) M1 \(v=2 \pi r / T\) M1 completion of algebra leading to \(B=2 \pi \mathrm{~m} / q T\) A1

Question 5(b)(ii)

5(b)(ii)

The particle is an alpha particle. The period of the circular motion is \(2.5 \mu \mathrm{~s}\). Calculate B.

Mediumstructured2 marks

Answer

\(B=\left(2 \pi \times 4 \times 1.66 \times 10^{-27}\right) /\left(2 \times 1.60 \times 10^{-19} \times 2.5 \times 10^{-6}\right)\) C1 \(=0.052 \mathrm{~T}\) A1

Question 5(b)(iii)

5(b)(iii)

A second alpha particle is in the same uniform field. It travels in a circle of radius 2 r. State and explain how the periods of the motion of the two particles compare.

Mediumstructured1 marks

Answer

either the same because T is independent of r or the same because B, q and m are unchanged or the same because both radius and speed have doubled B1

Question 5(b)(iv)

5(b)(iv)

The speed of the alpha particle in (b)(ii) is \(1.1 \times 10^{6} \mathrm{~ms}^{-1}\). An electric field is applied so that this particle now moves with constant velocity. Use your answer in (b)(ii) to calculate the electric field strength E. Give the unit with your answer. E= unit

Hardstructured2 marks

Answer

q E=B q v C1 \[ \begin{aligned} E =B V=0.052 \times 1.1 \times 10^{6} =5.7 \times 10^{4} \mathrm{~N} \mathrm{C}^{-1} \end{aligned} \] A1