IB Physics SL D.3 Motion in Electromagnetic Fields Question BankPractise IB Physics SL D.3 by analysing charged-particle motion in electric and magnetic fields with force, energy and trajectory evidence.SyllabusFirst assessment 2025CoursePhysics SLLevelSL
Exam pointsApply electric force and potential-energy relationships to charged-particle acceleration.Use magnetic-force direction and circular-motion reasoning for moving charges.Interpret trajectory, field-strength and energy data while stating vector assumptions.
D.3 Motion in electromagnetic fields question 1[Maximum number: 4]Two oppositely charged parallel plates are a distance 8.0 cm apart. The potential difference between the plates is 120 V . An alpha particle is placed on the positively charged plate and released from rest. Gravity is ignored.Question (a)(a)Show that the acceleration of the alpha particle is about 7×1010 ms−27 \times 10^{10} \mathrm{~ms}^{-2}7×1010 ms−2.[ 2 ]Show AnswerF=<eE=>3.2×10−19×1.5×103 OR 4.8×10−16 Na=<Fm=>4.8×10−164×1.67×10−27 OR 7.2×1010 ms−2\begin{aligned} F=<e E=>3.2 \times 10^{-19} \times 1.5 \times 10^{3} \text { OR } 4.8 \times 10^{-16} \mathrm{~N} a=<\frac{F}{m}=>\frac{4.8 \times 10^{-16}}{4 \times 1.67 \times 10^{-27}} \text { OR } 7.2 \times 10^{10} \mathrm{~ms}^{-2} \quad \end{aligned}F=<eE=>3.2×10−19×1.5×103 OR 4.8×10−16 Na=<mF=>4×1.67×10−274.8×10−16 OR 7.2×1010 ms−2Allow ECF from a).Award [1] if they use a charge of e and a mass of 2 u obtaining the right result.[2]Question (b)(b)A magnetic field directed into the plane of the page is now established between the plates. An alpha particle enters the region between the plates with a horizontal speed of 5.0×105 m s−15.0 \times 10^{5} \mathrm{~m} \mathrm{~s}^{-1}5.0×105 m s−1. The particle is not deflected.Calculate the magnitude of the magnetic field.[ 2 ]Show AnswerFm=Fe OR qvB=qE OR B=Fe/qvB=≪Ev=1.5×1035.0×105=>3.0×10−3T\begin{aligned} F_{\mathrm{m}} =F_{\mathrm{e}} \text { OR } q v B=q E \text { OR } B=F_{\mathrm{e}} / q v \quad B =\ll \frac{E}{v}=\frac{1.5 \times 10^{3}}{5.0 \times 10^{5}}=>3.0 \times 10^{-3} T \end{aligned}Fm=Fe OR qvB=qE OR B=Fe/qvB=≪vE=5.0×1051.5×103=>3.0×10−3TAward [2] if 3.0×10−33.0 \times 10^{-3}3.0×10−3 « T » is seen as the answer without working.Allow ECF from a) and b) i)[2]Add to Test
Question (a)(a)Show that the acceleration of the alpha particle is about 7×1010 ms−27 \times 10^{10} \mathrm{~ms}^{-2}7×1010 ms−2.[ 2 ]Show AnswerF=<eE=>3.2×10−19×1.5×103 OR 4.8×10−16 Na=<Fm=>4.8×10−164×1.67×10−27 OR 7.2×1010 ms−2\begin{aligned} F=<e E=>3.2 \times 10^{-19} \times 1.5 \times 10^{3} \text { OR } 4.8 \times 10^{-16} \mathrm{~N} a=<\frac{F}{m}=>\frac{4.8 \times 10^{-16}}{4 \times 1.67 \times 10^{-27}} \text { OR } 7.2 \times 10^{10} \mathrm{~ms}^{-2} \quad \end{aligned}F=<eE=>3.2×10−19×1.5×103 OR 4.8×10−16 Na=<mF=>4×1.67×10−274.8×10−16 OR 7.2×1010 ms−2Allow ECF from a).Award [1] if they use a charge of e and a mass of 2 u obtaining the right result.[2]
Question (b)(b)A magnetic field directed into the plane of the page is now established between the plates. An alpha particle enters the region between the plates with a horizontal speed of 5.0×105 m s−15.0 \times 10^{5} \mathrm{~m} \mathrm{~s}^{-1}5.0×105 m s−1. The particle is not deflected.Calculate the magnitude of the magnetic field.[ 2 ]Show AnswerFm=Fe OR qvB=qE OR B=Fe/qvB=≪Ev=1.5×1035.0×105=>3.0×10−3T\begin{aligned} F_{\mathrm{m}} =F_{\mathrm{e}} \text { OR } q v B=q E \text { OR } B=F_{\mathrm{e}} / q v \quad B =\ll \frac{E}{v}=\frac{1.5 \times 10^{3}}{5.0 \times 10^{5}}=>3.0 \times 10^{-3} T \end{aligned}Fm=Fe OR qvB=qE OR B=Fe/qvB=≪vE=5.0×1051.5×103=>3.0×10−3TAward [2] if 3.0×10−33.0 \times 10^{-3}3.0×10−3 « T » is seen as the answer without working.Allow ECF from a) and b) i)[2]
D.3 Motion in electromagnetic fields question 2[Maximum number: 4]Question (a)(a)A proton moves on a circular path in a region of uniform magnetic field of magnetic flux density B that is directed into the plane of the page.[ 4 ]Question (i)(i)On the diagram, draw an arrow to indicate the velocity of the proton at the position shown.[ 1 ]Show AnswerVertically down arrow from the protonIf more than one arrow is included thevelocity must be clearly labelled.[1]Question (ii)(ii)Show that the frequency of revolution of the proton is given by f=eB2πmpf=\frac{e B}{2 \pi m_{\mathrm{p}}}f=2πmpeB.[ 3 ]Show AnswerevB=mpv2Rf=1T=v2πR\begin{aligned} e v B=\frac{m_{\mathrm{p}} v^{2}}{R} f=\frac{1}{T}=\frac{v}{2 \pi R} \end{aligned}evB=Rmpv2f=T1=2πRvBoth q and e are acceptable for thecharge.[3]Algebra leading to required expression <=eB2πmp\boldsymbol{<}=\frac{e B}{2 \pi m_{\mathrm{p}}} \boldsymbol{} \boldsymbol{}<=2πmpeBAdd to Test
Question (a)(a)A proton moves on a circular path in a region of uniform magnetic field of magnetic flux density B that is directed into the plane of the page.[ 4 ]Question (i)(i)On the diagram, draw an arrow to indicate the velocity of the proton at the position shown.[ 1 ]Show AnswerVertically down arrow from the protonIf more than one arrow is included thevelocity must be clearly labelled.[1]Question (ii)(ii)Show that the frequency of revolution of the proton is given by f=eB2πmpf=\frac{e B}{2 \pi m_{\mathrm{p}}}f=2πmpeB.[ 3 ]Show AnswerevB=mpv2Rf=1T=v2πR\begin{aligned} e v B=\frac{m_{\mathrm{p}} v^{2}}{R} f=\frac{1}{T}=\frac{v}{2 \pi R} \end{aligned}evB=Rmpv2f=T1=2πRvBoth q and e are acceptable for thecharge.[3]Algebra leading to required expression <=eB2πmp\boldsymbol{<}=\frac{e B}{2 \pi m_{\mathrm{p}}} \boldsymbol{} \boldsymbol{}<=2πmpeB
Question (i)(i)On the diagram, draw an arrow to indicate the velocity of the proton at the position shown.[ 1 ]Show AnswerVertically down arrow from the protonIf more than one arrow is included thevelocity must be clearly labelled.[1]
Question (ii)(ii)Show that the frequency of revolution of the proton is given by f=eB2πmpf=\frac{e B}{2 \pi m_{\mathrm{p}}}f=2πmpeB.[ 3 ]Show AnswerevB=mpv2Rf=1T=v2πR\begin{aligned} e v B=\frac{m_{\mathrm{p}} v^{2}}{R} f=\frac{1}{T}=\frac{v}{2 \pi R} \end{aligned}evB=Rmpv2f=T1=2πRvBoth q and e are acceptable for thecharge.[3]Algebra leading to required expression <=eB2πmp\boldsymbol{<}=\frac{e B}{2 \pi m_{\mathrm{p}}} \boldsymbol{} \boldsymbol{}<=2πmpeB