IB Physics D.3: Motion in electromagnetic fields worked solutions

The field exerts a constant force eE at right angles to the plates, directed against the field lines for a negative charge, so towards the positive plate. A constant force at right angles to the initial velocity gives a parabola, like a ball thrown horizontally; nothing acts along the plates, so that velocity component stays constant. A is the magnetic-field path; C has the sign wrong.

a=\dfrac{eE}{m_p}=\dfrac{1.60\times10^{-19}\times3.0\times10^{4}}{1.673\times10^{-27}}=2.9\times10^{12}\,\mathrm{m\,s^{-2}}. A is the force in newtons, B mistakes the field strength for the acceleration, D uses the electron mass.

Check speed and angle first: A has v=0 and B has \sin0^\circ=0, so both feel no magnetic force however large the numbers. C: F=evB=1.60\times10^{-19}(1.0\times10^{5})(0.50)=8.0\times10^{-15}\,\mathrm{N}. D: F=2evB\sin30^\circ=3.2\times10^{-19}(1.5\times10^{5})(0.50)(0.50)=1.2\times10^{-14}\,\mathrm{N}, the largest.

For a positive charge the left-hand rule (field into the page, current to the right) gives a force up the page. The electron's charge is negative, so its conventional current points to the left and the force is reversed: down the page (B). C is the answer for a proton. The force is perpendicular to the field, which rules out D, and to the velocity, which rules out A.

(a) B=\dfrac{F}{IL}=\dfrac{9.0\times10^{-3}}{2.4\times0.15}=0.025\,\mathrm{T} (A1); convert mN to N first. (b) F=BIL\sin30^\circ=9.0\,\mathrm{mN}\times0.50=4.5\,\mathrm{mN} (A1). The angle is the one between the wire and the field, not between the wire and the page.

Left hand: first finger up (field), second finger north (current); the thumb points east. The force is at right angles to both the current and the field, so A and D are impossible; C comes from using the right hand.

The two forces are a Newton's third law pair: equal in size and opposite in direction, whatever the two currents are. \dfrac{F}{L}=\dfrac{\mu_0}{2\pi}\cdot\dfrac{I_1I_2}{r} contains the product I_1I_2, the same for both wires. Currents in opposite directions repel, so the force is away from the other wire.

The magnetic force never has a component along the velocity, so it does no work: the speed and the kinetic energy stay constant. The force, and so the acceleration v^2/r, keeps the same size. Momentum is a vector whose direction changes all the time, so it is not constant (C and D fail).

r=\dfrac{mv}{qB}: doubling v doubles r and halving B doubles it again, so the radius becomes 4r (A). B changes only one factor; D inverts both.

Sign first: both particles are positive, so they curve the same way (B fails). Then the radius: r=\dfrac{mv}{qB} with the same v and B gives r\propto\dfrac{m}{q}. The alpha particle has four times the mass and twice the charge, so its radius is about twice the proton's. D forgets the charge.

A charge at rest feels a force in an electric field and any mass feels one in a gravitational field, so A and C would make it move from rest. Only a magnetic field ignores a stationary charge, and its force at right angles to the velocity gives a circle at constant speed.

(a) The speed is constant, so T=\dfrac{2\pi r}{v}=\dfrac{2\pi(0.028)}{2.0\times10^{6}}=8.8\times10^{-8}\,\mathrm{s} (A1); use the circumference, not the diameter. (b) A quarter circle takes T/4=2.2\times10^{-8}\,\mathrm{s} (A1).

(a) E=\dfrac{V}{d}=\dfrac{90}{0.015}=6.0\times10^{3}\,\mathrm{V\,m^{-1}} (A1). (b) F=eE=1.60\times10^{-19}(6.0\times10^{3})=9.6\times10^{-16}\,\mathrm{N} (A1), upwards towards the positive plate, opposite to the field (A1). (c) mg=9.110\times10^{-31}(9.8)=8.9\times10^{-30}\,\mathrm{N} (M1); \dfrac{eE}{mg}=1.1\times10^{14}, so the weight changes the motion by about one part in 10^{14} (A1). Saying only that the electron is light does not earn the mark: compare the two forces. (d) a=\dfrac{eE}{m_e}=\dfrac{9.6\times10^{-16}}{9.110\times10^{-31}}=1.05\times10^{15}\,\mathrm{m\,s^{-2}} (A1).

(a) E=V/d=50/0.020=2.5\times10^{3}\,\mathrm{V\,m^{-1}} (M1); a=eE/m_p=1.60\times10^{-19}(2.5\times10^{3})/1.673\times10^{-27}=2.39\times10^{11}\,\mathrm{m\,s^{-2}} (A1). (b) The proton starts midway, so it hits a plate if it moves 1.0\,\mathrm{cm} sideways, not 2.0\,\mathrm{cm} (M1). Time to move 1.0\,\mathrm{cm}: t=\sqrt{2(0.010)/2.39\times10^{11}}=2.89\times10^{-7}\,\mathrm{s} (M1). Time to cross the plates: 0.060/1.0\times10^{5}=6.0\times10^{-7}\,\mathrm{s}, longer, so it hits a plate, 1.0\times10^{5}(2.89\times10^{-7})=2.9\,\mathrm{cm} from the entry (A1). Use the unchanged horizontal speed for the crossing time, never the total speed. (c) It must cover 6.0\,\mathrm{cm} in 2.89\times10^{-7}\,\mathrm{s} (M1): v=0.060/2.89\times10^{-7}=2.1\times10^{5}\,\mathrm{m\,s^{-1}} (A1).

An electron is pushed against the field direction, so here the force is directly against the velocity: the electron decelerates in a straight line, stops and is accelerated back. A is true for a proton. A parabola needs a force at an angle to the velocity, and a circle needs a magnetic field.

Two steps. Right-hand grip: with the current up the page, the wire's field to its right points into the page. Left-hand rule for the proton: field into the page, its motion up the page, so the force points to the left, towards the wire. Same result as for parallel currents in the same direction, which attract. C and D use the field direction as the force direction.

(a) The force is greatest when the wire is at right angles to the field: F=BIL=0.12(5.0)(0.40)=0.24\,\mathrm{N} (A1). (b) \sin\theta=\dfrac{F}{BIL}=\dfrac{0.12}{0.24}=0.50 (M1), \theta=30^\circ (A1). (c) Perpendicular to the wire and to the field, so perpendicular to the plane holding both (A1).

Take the top of the loop: the anticlockwise current there flows to the left. Left hand: field out of the page, current to the left, force towards the top of the page, away from the centre. The same holds at every point, so the forces pull outwards and cancel: no resultant force and no turning, but the loop stretches. Every force lies in the plane of the loop, so D is impossible.

(a) Each cable sits in the magnetic field of the other and carries a current across it, so it feels a force BIL (M1); for currents in the same direction the forces pull the cables together: they attract (A1). (b) \dfrac{F}{L}=\dfrac{\mu_0}{2\pi}\cdot\dfrac{I_1I_2}{r}=\dfrac{4\pi\times10^{-7}(600)(600)}{2\pi(0.30)}=0.24\,\mathrm{N\,m^{-1}} (A1). (c) \dfrac{F}{L}=\dfrac{2\times10^{-7}(1200)(600)}{0.45}=0.32\,\mathrm{N\,m^{-1}} (M1 for both changes, A1). (d) The forces form a Newton's third law pair: the same size on both cables, in opposite directions, even though one current is larger (A1).

(a) The magnetic force is perpendicular to the velocity at every instant, so it does no work and the speed stays constant (A1). At constant speed the force has constant size and always points towards the same point, so it acts as the centripetal force of a circle (A1). (b) evB=\dfrac{m_pv^2}{r} gives v=\dfrac{eBr}{m_p}=\dfrac{1.60\times10^{-19}(0.060)(0.25)}{1.673\times10^{-27}} (M1) =1.43\times10^{6}\,\mathrm{m\,s^{-1}} (A1). (c) E_k=\tfrac12m_pv^2=1.72\times10^{-15}\,\mathrm{J} (A1); 1.72\times10^{-15}/1.60\times10^{-19}=1.08\times10^{4}\,\mathrm{eV} (A1).

(a) The field does work qV with q=2e (M1): \tfrac12mv^2=2(1.60\times10^{-19})(2000)=6.4\times10^{-16}\,\mathrm{J}, v=\sqrt{2(6.4\times10^{-16})/6.64\times10^{-27}}=4.39\times10^{5}\,\mathrm{m\,s^{-1}} (A1). (b) r=\dfrac{mv}{qB}=\dfrac{6.64\times10^{-27}(4.39\times10^{5})}{3.20\times10^{-19}(0.25)} (M1) =3.6\times10^{-2}\,\mathrm{m} (A1). (c) For a proton v=\sqrt{2eV/m_p}=6.19\times10^{5}\,\mathrm{m\,s^{-1}} (M1) and r=\dfrac{m_pv}{eB}=2.6\times10^{-2}\,\mathrm{m} (A1). The particles share the p.d., not the speed, so r\propto\sqrt{m/q}, not m/q.

(a) X: the force acts on a charge at rest but not on a neutral particle, so it is electric (A1). Y: no force on a stationary charge, but a moving charge is pushed at right angles to its velocity, so it is magnetic (A1). Z: neutral and charged particles accelerate equally, independent of charge and mass, as in free fall: gravitational (A1).

(a) r=\dfrac{m_ev}{eB}=\dfrac{9.110\times10^{-31}(3.5\times10^{6})}{1.60\times10^{-19}(1.2\times10^{-3})} (M1) =1.66\times10^{-2}\,\mathrm{m} (A1); 1.2 mT is 1.2\times10^{-3}\,\mathrm{T}. (b) T=\dfrac{2\pi r}{v} (M1) and r=\dfrac{m_ev}{eB}, so T=\dfrac{2\pi m_e}{eB}: v cancels (A1). A faster electron goes round a larger circle in the same time. (c) T=\dfrac{2\pi(9.110\times10^{-31})}{1.60\times10^{-19}(1.2\times10^{-3})}=2.98\times10^{-8}\,\mathrm{s} (M1); a quarter turn takes T/4=7.5\times10^{-9}\,\mathrm{s} (A1).

(a) The electrons move to the right with the rod. For a positive charge the force would be up the page; electrons are negative, so the force on them is down the rod (A1). (b) Electrons collect at the lower end, leaving the upper end positive; this charge sets up an electric field along the rod whose force on each electron is upwards (M1). Electrons stop collecting when eE=evB (A1). (c) E=vB=4.0(0.20)=0.80\,\mathrm{V\,m^{-1}} (M1); V=EL=0.80(0.50)=0.40\,\mathrm{V}, that is V=BvL (A1).

(a) The electric force is downwards; qvB acts upwards. With zero resultant force, qE=qvB (M1), so v=E/B=1.8\times10^{4}/0.060=3.0\times10^{5}\,\mathrm{m\,s^{-1}} (A1). (b) At greater speed the magnetic force increases while the electric force is unchanged; the ion initially bends upwards (A1). The selected speed is independent of mass and charge magnitude.

Use \dfrac{F}{L}=\dfrac{\mu_0}{2\pi}\cdot\dfrac{I_1I_2}{r}=\dfrac{2\times10^{-7}I_1I_2}{r} for each pair. (a) P on Q: same direction, attraction towards P, 2\times10^{-7}(4.0)(2.0)/0.050=3.2\times10^{-5}\,\mathrm{N\,m^{-1}} (M1). R on Q: opposite, repulsion away from R, which is also towards P, 2\times10^{-7}(6.0)(2.0)/0.050=4.8\times10^{-5}\,\mathrm{N\,m^{-1}} (M1). Total 8.0\times10^{-5}\,\mathrm{N\,m^{-1}} towards P (A1). (b) Q on P: 3.2\times10^{-5} towards Q; R on P (distance 0.10\,\mathrm{m}): 2\times10^{-7}(4.0)(6.0)/0.10=4.8\times10^{-5} away from R (M1). Resultant 1.6\times10^{-5}\,\mathrm{N\,m^{-1}} away from Q (A1). (c) Taking the direction from P to R as positive: -8.0-1.6+9.6=0 (A1). Each pair of wires exerts equal and opposite forces on each other, so the internal forces of the set of three must add to zero (A1).

(a) a=\dfrac{eV}{m_ed}=\dfrac{1.60\times10^{-19}(120)}{9.110\times10^{-31}(0.016)}=1.32\times10^{15}\,\mathrm{m\,s^{-2}}; t=0.040/2.5\times10^{7}=1.6\times10^{-9}\,\mathrm{s} (M1); y=\tfrac12at^2=1.7\times10^{-3}\,\mathrm{m}, less than half the gap, 8.0\,\mathrm{mm} (A1). (b) v_y=at=2.11\times10^{6}\,\mathrm{m\,s^{-1}} (M1); \tan\theta=v_y/v_x=0.0843, \theta=4.8^\circ (A1). Use the velocity components, not the displacements. (c) Beyond the plates there is no field and the path is straight: extra deflection 0.20\tan\theta=1.69\times10^{-2}\,\mathrm{m} (M1); total 1.7\,\mathrm{mm}+16.9\,\mathrm{mm}=18.5\,\mathrm{mm}\approx1.9\,\mathrm{cm} (A1). (d) Both parts are proportional to 1/v^2, so the deflection falls to one quarter (A1).

(a) Energy: eV=\tfrac12m_ev^2 (M1). Circle: evB=\dfrac{m_ev^2}{r}, so v=\dfrac{eBr}{m_e}; substituting, eV=\tfrac12m_e\dfrac{e^2B^2r^2}{m_e^2}, which gives V=\dfrac{eB^2}{2m_e}r^2 (A1). (b) r^2 / m^2: 9.0,16,25,36,49\times10^{-4}; the points lie on a line through the origin (M1); gradient =\dfrac{276-51}{(49-9.0)\times10^{-4}}=5.63\times10^{4}\,\mathrm{V\,m^{-2}} (A1). Plotting V against r gives a curve. (c) \dfrac{e}{m_e}=\dfrac{2\times\text{gradient}}{B^2} (M1) =\dfrac{2(5.63\times10^{4})}{(0.80\times10^{-3})^2}=1.76\times10^{11}\,\mathrm{C\,kg^{-1}} (A1); divide by B^2, not B. (d) e and m_e enter only as e/m_e: a particle with twice the charge and twice the mass would follow the same path (A1).

(a) Take a point where the track moves to the right. For a positive charge, field out of the page and motion to the right give a force towards the bottom of the page, a clockwise turn (M1). The track turns the other way, so the charge is negative (A1). (b) v=\dfrac{eBr}{m_e}=\dfrac{1.60\times10^{-19}(2.0\times10^{-3})(0.040)}{9.110\times10^{-31}} (M1) =1.4\times10^{7}\,\mathrm{m\,s^{-1}} (A1), about 5\% of c, so r=mv/(eB) applies. (c) The magnetic force is at right angles to the velocity and does no work, so it cannot slow the electron (M1). The electron loses kinetic energy in collisions with the atoms of the detector; its speed falls, and r\propto v in a fixed field (A1). (d) v\propto r, so E_k\propto r^2: (2.0/4.0)^2=0.25 (A1). The track is planar, not a helix.

(a) The 8.0\,\mathrm{cm} sides carry current parallel or antiparallel to the field: \sin\theta=0, no force (A1). Each 5.0\,\mathrm{cm} side: F=BIL=0.15(3.0)(0.050)=2.25\times10^{-2}\,\mathrm{N} (A1). Clockwise current flows down the right side and up the left side; the left-hand rule gives out of the page on the right side and into the page on the left side (A1). (b) The two forces are equal in size and opposite in direction, so the resultant is zero (A1); they act along different lines, so the loop turns about the axis, its right side coming out of the page (A1). (c) The side that came towards you still carries current down the page; with the field to the right, its force points further towards you, away from the centre. Every side is pushed outwards in the plane of the loop (A1); the forces cancel and have no turning effect, so they only stretch the loop (A1).

(a) B.5: the work done on each electron is eV=1.60\times10^{-19}(1800)=2.88\times10^{-16}\,\mathrm{J} (A1). (b) A.3: \tfrac12m_ev^2=2.88\times10^{-16} (M1), v=\sqrt{2(2.88\times10^{-16})/9.110\times10^{-31}}=2.51\times10^{7}\,\mathrm{m\,s^{-1}} (A1), under a tenth of c. (c) B.5: n=I/e=0.40\times10^{-3}/1.60\times10^{-19}=2.5\times10^{15} each second (A1). (d) r=\dfrac{m_ev}{eB}=\dfrac{9.110\times10^{-31}(2.51\times10^{7})}{1.60\times10^{-19}(1.5\times10^{-3})} (M1) =9.5\times10^{-2}\,\mathrm{m} (A1). (e) P=IV=0.40\times10^{-3}(1800)=0.72\,\mathrm{W} (A1); the magnetic force never has a component along the velocity, so it does no work and the speed of every electron is unchanged (A1).

(a) A.2: the two springs share the weight, 2kx=mg (M1): k=\dfrac{0.025(9.8)}{2(0.020)}=6.1\,\mathrm{N\,m^{-1}} (A1). (b) The magnetic force must equal the weight: BIL=mg (M1), I=\dfrac{0.245}{0.30(0.20)}=4.1\,\mathrm{A} (A1). (c) Reversed, the magnetic force acts downwards and adds to the weight: the springs now hold 2mg (M1), so each extends 2(2.0)=4.0\,\mathrm{cm} (A1). (d) The force must be upwards. Left hand: field north, thumb up, second finger points east: the current flows from west to east (A1).

(a) E.1: nucleon number 22 minus 10 protons: 12 neutrons (A1). (b) m=20(1.661\times10^{-27})=3.32\times10^{-26}\,\mathrm{kg}; \tfrac12mv^2=eV=8.0\times10^{-17}\,\mathrm{J} (M1); v=\sqrt{2(8.0\times10^{-17})/3.32\times10^{-26}}=6.94\times10^{4}\,\mathrm{m\,s^{-1}} (A1). (c) r=\dfrac{mv}{eB}=\dfrac{3.32\times10^{-26}(6.94\times10^{4})}{1.60\times10^{-19}(0.20)} (M1) =7.2\times10^{-2}\,\mathrm{m} (A1). (d) Same charge and same p.d., so the same kinetic energy: r=\dfrac{\sqrt{2meV}}{eB}\propto\sqrt m, r_{22}=7.2\times10^{-2}\sqrt{22/20}=7.6\times10^{-2}\,\mathrm{m} (A1). Isotopes have the same charge but different numbers of neutrons, so different masses, and the heavier ions curve less (A1).

(a) E=V/d=675/0.15=4.5\times10^{3}\,\mathrm{V\,m^{-1}} (A1); F=qE=1.8\times10^{-11}\,\mathrm{N}, upwards since the lower plate is positive (A1). (b) A.2: the weight mg=1.96\times10^{-11}\,\mathrm{N} acts down, against the electric force, so they subtract: resultant 1.6\times10^{-12}\,\mathrm{N} downwards (M1); a=F/m=0.80\,\mathrm{m\,s^{-2}} (A1). Here the weight is NOT negligible. (c) A.1: time between the plates 0.20/0.50=0.40\,\mathrm{s}; y=\tfrac12at^2=\tfrac12(0.80)(0.40)^2=0.064\,\mathrm{m} (A1), less than half the gap, 0.075\,\mathrm{m}, so it leaves the plates (A1). (d) Straight through when qE=mg: E=mg/q=4.9\times10^{3}\,\mathrm{V\,m^{-1}}, V=Ed=735\,\mathrm{V} (A1).