Fields and their consequences (A-level only)

3.7 Fields and their consequences (A-level only) — coverage pack

18 specification leaves · notes, questions, answers and worked methods

3.7.1 · Fields

  • A force field is a region in which a body experiences a non-contact force; gravitational fields act on mass and electric fields act on charge.
  • Field strength is a vector. Its direction is the direction of the force on a small positive test charge, or on a test mass for gravity.
  • Gravitational and electrostatic forces both follow inverse-square laws and can be represented by field lines, potentials and equipotential surfaces.
  • Masses always attract, whereas like charges repel and unlike charges attract. A common error is to infer a field direction from the sign of potential alone.

Tier 1 · Easy

  1. 1. State one difference between a gravitational force field and an electric force field.[1 mark]

    Answer

    • A gravitational field always produces attraction between masses, whereas an electric field can produce attraction or repulsion between charges.

    Method: Award the mark for a valid contrast: gravitational interaction between masses is always attractive, but the electric interaction depends on the signs of the charges and may be attractive or repulsive.

Tier 2 · Standard

  1. 1. A small positive charge and a small mass are placed separately at the same point near a negatively charged metal sphere and a planet. Describe the direction of each force and explain how a field-line diagram shows field strength.[3 marks]

    Answer

    • Both forces are towards the source; the electric force is towards the negative sphere and the gravitational force is towards the planet. A greater field-line density represents a stronger field.

    Method: A positive test charge is attracted to the negative sphere, so its force is directed towards the sphere. A test mass is attracted to the planet, so its force is directed towards the planet. In either diagram, closer field lines indicate a larger field-strength magnitude; the arrows give the vector direction.

Tier 3 · Hard

  1. 1. Two identical positive point charges are fixed a short distance apart. Discuss the electric field and electric potential at the midpoint, and describe the required relationship between field lines and equipotential surfaces near the charges.[5 marks]

    Answer

    • At the midpoint the two electric-field vectors cancel, so the resultant field is zero, but the two positive potentials add, so the potential is positive and non-zero. Field lines leave each positive charge, never cross, and meet equipotential surfaces at right angles; moving along an equipotential requires no work.

    Method: Treat electric field as a vector: the equal fields at the midpoint point in opposite directions and cancel. Treat potential as a scalar: each positive charge contributes a positive potential, so the contributions add rather than cancel. Field lines point away from positive charges and cannot cross because the field has one direction at a point. Since no work is done along an equipotential, the electric field must be perpendicular to every equipotential surface.

3.7.2.1 · Newton's law

  • Newton's law of gravitation gives the magnitude between point masses: F=Gm1m2r2F=\dfrac{Gm_1m_2}{r^2}, where rr is their centre-to-centre separation.
  • Gravity is universal and attractive: the two bodies experience equal-magnitude, opposite-direction forces, consistent with Newton's third law.
  • For a spherically symmetric body, its mass may be treated as concentrated at its centre when calculating the external force.
  • Use G=6.67×10−11 N m2 kg−2G=6.67\times10^{-11}\,\text{N m}^2\text{ kg}^{-2}. A common error is to use surface gap or diameter instead of centre-to-centre distance.

Tier 1 · Easy

  1. 1. Calculate the gravitational force between masses 6.0×1022 kg6.0\times10^{22}\,\text{kg} and 4.0×1020 kg4.0\times10^{20}\,\text{kg} whose centres are 3.0×107 m3.0\times10^7\,\text{m} apart.[2 marks]

    Answer

    • 1.8×1018 N1.8\times10^{18}\,\text{N}, attractive

    Method: Use F=Gm1m2r2F=\dfrac{Gm_1m_2}{r^2}. Hence F=(6.67×10−11)(6.0×1022)(4.0×1020)(3.0×107)2=1.78×1018 NF=\dfrac{(6.67\times10^{-11})(6.0\times10^{22})(4.0\times10^{20})}{(3.0\times10^7)^2}=1.78\times10^{18}\,\text{N}. To two significant figures this is 1.8×1018 N1.8\times10^{18}\,\text{N}, and gravity makes the force attractive.

Tier 2 · Standard

  1. 1. An 850 kg850\,\text{kg} satellite is 7.20×106 m7.20\times10^6\,\text{m} from the centre of a planet of mass 5.97×1024 kg5.97\times10^{24}\,\text{kg}. Calculate the gravitational force on the satellite and its acceleration. Explain why the acceleration would be unchanged for a satellite of different mass at the same point.[4 marks]

    Answer

    • Force 6.53×103 N6.53\times10^3\,\text{N}; acceleration 7.68 m s−27.68\,\text{m s}^{-2}, independent of satellite mass.

    Method: The force is F=GMmr2=(6.67×10−11)(5.97×1024)(850)(7.20×106)2=6.53×103 NF=\dfrac{GMm}{r^2}=\dfrac{(6.67\times10^{-11})(5.97\times10^{24})(850)}{(7.20\times10^6)^2}=6.53\times10^3\,\text{N}. Then a=F/m=(6.53×103)/850=7.68 m s−2a=F/m=(6.53\times10^3)/850=7.68\,\text{m s}^{-2}. Substituting Newton's law into a=F/ma=F/m cancels the satellite mass: a=GM/r2a=GM/r^2.

Tier 3 · Hard

  1. 1. Two approximately spherical bodies have masses 4.0×1024 kg4.0\times10^{24}\,\text{kg} and 7.5×1023 kg7.5\times10^{23}\,\text{kg}. Their mutual gravitational force is 8.0×1020 N8.0\times10^{20}\,\text{N}. Determine their centre-to-centre separation and the new force if that separation increases by 20%20\%.[5 marks]

    Answer

    • Separation 5.0×108 m5.0\times10^8\,\text{m}; new force 5.6×1020 N5.6\times10^{20}\,\text{N}.

    Method: Rearrange Newton's law to r=Gm1m2/Fr=\sqrt{Gm_1m_2/F}. Thus r=(6.67×10−11)(4.0×1024)(7.5×1023)/(8.0×1020)=5.00×108 mr=\sqrt{(6.67\times10^{-11})(4.0\times10^{24})(7.5\times10^{23})/(8.0\times10^{20})}=5.00\times10^8\,\text{m}. The new separation is 1.20r1.20r. Since F∝1/r2F\propto1/r^2, Fnew=8.0×1020/(1.20)2=5.56×1020 NF_{\text{new}}=8.0\times10^{20}/(1.20)^2=5.56\times10^{20}\,\text{N}, or 5.6×1020 N5.6\times10^{20}\,\text{N} to two significant figures.

3.7.2.2 · Gravitational field strength

  • Gravitational field strength is force per unit mass: g=F/mg=F/m. Its unit is N kg−1\text{N kg}^{-1}, equivalent to m s−2\text{m s}^{-2}.
  • Outside a spherical mass, the radial-field magnitude is g=GM/r2g=GM/r^2 and the vector points towards the centre of mass.
  • Radial field lines point inward and become less dense with increasing rr, showing the inverse-square decrease in field strength.
  • Use distance from the centre, not altitude above the surface. A common error is to omit the inward direction when a vector answer is required.

Tier 1 · Easy

  1. 1. Define gravitational field strength at a point.[1 mark]

    Answer

    • Gravitational field strength is the gravitational force per unit mass on a small test mass at that point.

    Method: State both the force-per-unit-mass ratio and the test mass at the point: g=F/mg=F/m. Gravitational field strength is a vector directed as the force on that test mass.

Tier 2 · Standard

  1. 1. Determine the gravitational field strength 2.40×107 m2.40\times10^7\,\text{m} from the centre of a planet of mass 4.80×1024 kg4.80\times10^{24}\,\text{kg}.[3 marks]

    Answer

    • 0.556 N kg−10.556\,\text{N kg}^{-1} towards the planet

    Method: Use g=GM/r2g=GM/r^2. Substitution gives g=(6.67×10−11)(4.80×1024)/(2.40×107)2=0.5558 N kg−1g=(6.67\times10^{-11})(4.80\times10^{24})/(2.40\times10^7)^2=0.5558\,\text{N kg}^{-1}. Therefore g=0.556 N kg−1g=0.556\,\text{N kg}^{-1} to three significant figures, directed radially towards the planet.

Tier 3 · Hard

  1. 1. A planet has mass 6.40×1024 kg6.40\times10^{24}\,\text{kg} and radius 7.00×106 m7.00\times10^6\,\text{m}. Calculate the altitude above its surface where the gravitational field strength is 2.40 N kg−12.40\,\text{N kg}^{-1}.[5 marks]

    Answer

    • 6.34×106 m6.34\times10^6\,\text{m} above the surface

    Method: From g=GM/r2g=GM/r^2, the distance from the centre is r=GM/gr=\sqrt{GM/g}. Hence r=(6.67×10−11)(6.40×1024)/2.40=1.3337×107 mr=\sqrt{(6.67\times10^{-11})(6.40\times10^{24})/2.40}=1.3337\times10^7\,\text{m}. Altitude is not this radial distance: h=r−R=1.3337×107−7.00×106=6.34×106 mh=r-R=1.3337\times10^7-7.00\times10^6=6.34\times10^6\,\text{m} to three significant figures.

3.7.2.3 · Gravitational potential

  • Gravitational potential is work done per unit mass in bringing a small mass from infinity to a point; the zero is defined at infinity.
  • For a spherical mass, Vg=−GM/rV_g=-GM/r. The negative sign shows that work must be supplied to move a mass from a bound position to infinity.
  • For movement between two points, the external work done in a slow transfer is ΔW=mΔVg=m(Vfinal−Vinitial)\Delta W=m\Delta V_g=m(V_{\text{final}}-V_{\text{initial}}).
  • Equipotentials are perpendicular to field lines, and no work is done along one. Also g=−ΔVg/Δrg=-\Delta V_g/\Delta r; a common error is to treat zero field as necessarily zero potential.

Tier 1 · Easy

  1. 1. State why gravitational potential near an isolated planet is negative when potential is defined as zero at infinity.[1 mark]

    Answer

    • The gravitational force attracts a mass, so positive external work is required to move the mass from the point to infinity; the point therefore has lower potential than infinity.

    Method: Infinity is assigned Vg=0V_g=0. A mass at finite distance is gravitationally bound and energy must be added to remove it to infinity, so its potential energy per unit mass, and hence VgV_g, is negative.

Tier 2 · Standard

  1. 1. A 320 kg320\,\text{kg} probe moves slowly from 9.0×106 m9.0\times10^6\,\text{m} to 1.8×107 m1.8\times10^7\,\text{m} from the centre of a planet of mass 7.2×1024 kg7.2\times10^{24}\,\text{kg}. Calculate the change in gravitational potential and the work done by the external force.[4 marks]

    Answer

    • ΔVg=+2.7×107 J kg−1\Delta V_g=+2.7\times10^7\,\text{J kg}^{-1}; external work +8.5×109 J+8.5\times10^9\,\text{J}.

    Method: The initial potential is Vi=−GM/ri=−(6.67×10−11)(7.2×1024)/(9.0×106)=−5.34×107 J kg−1V_i=-GM/r_i=-(6.67\times10^{-11})(7.2\times10^{24})/(9.0\times10^6)=-5.34\times10^7\,\text{J kg}^{-1}. The final potential is Vf=−2.67×107 J kg−1V_f=-2.67\times10^7\,\text{J kg}^{-1}. Therefore ΔVg=Vf−Vi=+2.67×107 J kg−1\Delta V_g=V_f-V_i=+2.67\times10^7\,\text{J kg}^{-1} and ΔW=mΔVg=320(2.67×107)=+8.54×109 J\Delta W=m\Delta V_g=320(2.67\times10^7)=+8.54\times10^9\,\text{J}.

Tier 3 · Hard

  1. 1. A 450 kg450\,\text{kg} craft is moved slowly from radius 8.0×106 m8.0\times10^6\,\text{m} to radius 2.4×107 m2.4\times10^7\,\text{m} from a body of mass 5.5×1024 kg5.5\times10^{24}\,\text{kg}. Determine the external work done. The craft is then released from rest at the outer point; calculate its speed when it returns to the inner point.[6 marks]

    Answer

    • External work +1.38×1010 J+1.38\times10^{10}\,\text{J}; return speed 7.82×103 m s−17.82\times10^3\,\text{m s}^{-1}.

    Method: At the inner point VA=−GM/rA=−4.5856×107 J kg−1V_A=-GM/r_A=-4.5856\times10^7\,\text{J kg}^{-1}; at the outer point VB=−GM/rB=−1.5285×107 J kg−1V_B=-GM/r_B=-1.5285\times10^7\,\text{J kg}^{-1}. For the outward move, ΔVg=VB−VA=+3.0571×107 J kg−1\Delta V_g=V_B-V_A=+3.0571\times10^7\,\text{J kg}^{-1}, so external work is mΔVg=450(3.0571×107)=+1.38×1010 Jm\Delta V_g=450(3.0571\times10^7)=+1.38\times10^{10}\,\text{J}. On return, ΔVg=VA−VB=−3.0571×107 J kg−1\Delta V_g=V_A-V_B=-3.0571\times10^7\,\text{J kg}^{-1} and conservation of energy gives 12mv2=−mΔVg\tfrac12mv^2=-m\Delta V_g. Thus v=−2ΔVg=7.82×103 m s−1v=\sqrt{-2\Delta V_g}=7.82\times10^3\,\text{m s}^{-1}.

3.7.2.4 · Orbits of planets and satellites

  • For a circular orbit, gravity supplies the centripetal force: GMm/r2=mv2/rGMm/r^2=mv^2/r, so v=GM/rv=\sqrt{GM/r}.
  • Using v=2πr/Tv=2\pi r/T with Newton's law gives T2=4π2r3/(GM)T^2=4\pi^2r^3/(GM) and hence T2∝r3T^2\propto r^3 for one central mass.
  • The total energy of a circular orbit is E=−GMm/(2r)E=-GMm/(2r); moving to a larger circular orbit makes the total energy less negative.
  • A geostationary satellite has a 2424-hour period, travels west-to-east above the equator and stays over one point. A common error is to use altitude instead of orbital radius in equations.
  • Escape speed follows from setting total energy at infinity to zero: ve=2GM/rv_e=\sqrt{2GM/r}.

Tier 1 · Easy

  1. 1. A satellite completes a circular orbit of radius 7.5×106 m7.5\times10^6\,\text{m} in 6.0×103 s6.0\times10^3\,\text{s}. Calculate its orbital speed.[2 marks]

    Answer

    • 7.9×103 m s−17.9\times10^3\,\text{m s}^{-1}

    Method: The distance travelled in one orbit is 2πr2\pi r. Therefore v=2πr/T=2π(7.5×106)/(6.0×103)=7.85×103 m s−1v=2\pi r/T=2\pi(7.5\times10^6)/(6.0\times10^3)=7.85\times10^3\,\text{m s}^{-1}, which is 7.9×103 m s−17.9\times10^3\,\text{m s}^{-1} to two significant figures.

Tier 2 · Standard

  1. 1. Starting from Newton's law of gravitation, show that T2∝r3T^2\propto r^3 for circular orbits around one planet. Hence determine the factor by which the period changes when orbital radius is multiplied by 1.601.60.[4 marks]

    Answer

    • The period is multiplied by 2.022.02.

    Method: Newton's law provides the centripetal force: GMm/r2=mv2/rGMm/r^2=mv^2/r. With v=2πr/Tv=2\pi r/T, this becomes GMm/r2=m(4π2r2/T2)/rGMm/r^2=m(4\pi^2r^2/T^2)/r. Cancelling mm and rearranging gives T2=4π2r3/(GM)T^2=4\pi^2r^3/(GM), so T2∝r3T^2\propto r^3. Therefore Tnew/Told=(1.60)3/2=2.02T_{\text{new}}/T_{\text{old}}=(1.60)^{3/2}=2.02.

Tier 3 · Hard

  1. 1. A planet of mass 6.0×1024 kg6.0\times10^{24}\,\text{kg} and radius 7.0×106 m7.0\times10^6\,\text{m} rotates once every 30 h30\,\text{h}. Determine the radius and altitude of a synchronous circular orbit. Calculate the total energy of a 500 kg500\,\text{kg} satellite in that orbit.[6 marks]

    Answer

    • Orbital radius 4.91×107 m4.91\times10^7\,\text{m}; altitude 4.21×107 m4.21\times10^7\,\text{m}; total energy −2.04×109 J-2.04\times10^9\,\text{J}.

    Method: Newton's gravitational force supplies centripetal force, giving T2=4π2r3/(GM)T^2=4\pi^2r^3/(GM). Convert T=30×3600=1.08×105 sT=30\times3600=1.08\times10^5\,\text{s}. Then r=[GMT2/(4π2)]1/3=4.908×107 mr=[GMT^2/(4\pi^2)]^{1/3}=4.908\times10^7\,\text{m}. The altitude is r−R=4.908×107−7.0×106=4.21×107 mr-R=4.908\times10^7-7.0\times10^6=4.21\times10^7\,\text{m}. For a circular orbit, E=−GMm/(2r)=−(6.67×10−11)(6.0×1024)(500)/(2×4.908×107)=−2.04×109 JE=-GMm/(2r)=-(6.67\times10^{-11})(6.0\times10^{24})(500)/(2\times4.908\times10^7)=-2.04\times10^9\,\text{J}.

3.7.3.1 · Coulomb's law

  • For point charges in a vacuum, F=14πε0∣Q1Q2∣r2F=\dfrac{1}{4\pi\varepsilon_0}\dfrac{|Q_1Q_2|}{r^2}, where rr is the separation and 1/(4πε0)=8.99×109 N m2 C−21/(4\pi\varepsilon_0)=8.99\times10^9\,\text{N m}^2\text{ C}^{-2}.
  • Like charges repel and unlike charges attract. Coulomb's law gives a magnitude, so direction must be determined separately from the charge signs and geometry.
  • Air may normally be treated as a vacuum, and the external field of a charged sphere may be modelled as if its charge were at the centre.
  • For several charges, calculate each force vector before adding components. A common error is to add force magnitudes when the forces are not collinear.

Tier 1 · Easy

  1. 1. Point charges +3.0 nC+3.0\,\text{nC} and −8.0 nC-8.0\,\text{nC} are 0.120 m0.120\,\text{m} apart in air. Calculate the force between them.[2 marks]

    Answer

    • 1.50×10−5 N1.50\times10^{-5}\,\text{N}, attractive

    Method: Treat air as a vacuum and use F=k∣Q1Q2∣/r2F=k|Q_1Q_2|/r^2. Thus F=(8.99×109)(3.0×10−9)(8.0×10−9)/(0.120)2=1.50×10−5 NF=(8.99\times10^9)(3.0\times10^{-9})(8.0\times10^{-9})/(0.120)^2=1.50\times10^{-5}\,\text{N}. The opposite signs make the force attractive.

Tier 2 · Standard

  1. 1. Determine the ratio of the electrostatic force to the gravitational force between a proton and an electron. Use mp=1.67×10−27 kgm_p=1.67\times10^{-27}\,\text{kg} and me=9.11×10−31 kgm_e=9.11\times10^{-31}\,\text{kg}, and explain why their separation is not needed.[4 marks]

    Answer

    • FE/FG=2.27×1039F_E/F_G=2.27\times10^{39}

    Method: The force magnitudes are FE=ke2/r2F_E=ke^2/r^2 and FG=Gmpme/r2F_G=Gm_pm_e/r^2. Dividing cancels the common r2r^2: FE/FG=ke2/(Gmpme)F_E/F_G=ke^2/(Gm_pm_e). Substitution gives (8.99×109)(1.60×10−19)2/[(6.67×10−11)(1.67×10−27)(9.11×10−31)]=2.27×1039(8.99\times10^9)(1.60\times10^{-19})^2/[(6.67\times10^{-11})(1.67\times10^{-27})(9.11\times10^{-31})]=2.27\times10^{39}.

Tier 3 · Hard

  1. 1. A charge +2.0 nC+2.0\,\text{nC} is at the origin. A charge +5.0 nC+5.0\,\text{nC} is at (0.300 m,0)(0.300\,\text{m},0) and a charge −4.0 nC-4.0\,\text{nC} is at (0,0.400 m)(0,0.400\,\text{m}). Determine the magnitude and direction of the resultant electrostatic force on the charge at the origin.[6 marks]

    Answer

    • 1.10×10−6 N1.10\times10^{-6}\,\text{N}, 24.2∘24.2^\circ above the negative xx-axis.

    Method: The +5.0 nC+5.0\,\text{nC} charge repels the origin charge in the negative xx direction: Fx=−(8.99×109)(2.0×10−9)(5.0×10−9)/(0.300)2=−9.99×10−7 NF_x=-(8.99\times10^9)(2.0\times10^{-9})(5.0\times10^{-9})/(0.300)^2=-9.99\times10^{-7}\,\text{N}. The negative charge attracts it in the positive yy direction: Fy=(8.99×109)(2.0×10−9)(4.0×10−9)/(0.400)2=4.50×10−7 NF_y=(8.99\times10^9)(2.0\times10^{-9})(4.0\times10^{-9})/(0.400)^2=4.50\times10^{-7}\,\text{N}. Hence F=Fx2+Fy2=1.10×10−6 NF=\sqrt{F_x^2+F_y^2}=1.10\times10^{-6}\,\text{N} and θ=tan⁡−1(∣Fy/Fx∣)=24.2∘\theta=\tan^{-1}(|F_y/F_x|)=24.2^\circ above the negative xx-axis.

3.7.3.2 · Electric field strength

  • Electric field strength is force per unit positive charge: E=F/QE=F/Q, with unit N C−1\text{N C}^{-1} or V m−1\text{V m}^{-1}.
  • For a point charge, E=Q/(4πε0r2)E=Q/(4\pi\varepsilon_0r^2) in magnitude; field lines point away from positive charge and towards negative charge.
  • Between parallel plates the field is approximately uniform and E=V/dE=V/d. From work, Fd=QΔVFd=Q\Delta V gives the same result.
  • A charged particle entering a uniform field perpendicular to its initial velocity has constant acceleration parallel to the field and constant velocity perpendicular to it, producing a parabolic path.
  • A common error is to reverse the field direction for an electron: the field follows force on positive charge, while an electron's force is opposite to EE.

Tier 1 · Easy

  1. 1. Define electric field strength at a point.[1 mark]

    Answer

    • Electric field strength is the force per unit positive test charge at that point.

    Method: Use E=F/QE=F/Q and specify a positive test charge so that the definition also fixes the direction of the field vector.

Tier 2 · Standard

  1. 1. Parallel plates have a potential difference of 1.80 kV1.80\,\text{kV} and separation 45.0 mm45.0\,\text{mm}. Calculate the uniform electric field strength, the force magnitude on an electron, and its acceleration magnitude.[4 marks]

    Answer

    • E=4.00×104 V m−1E=4.00\times10^4\,\text{V m}^{-1}; F=6.40×10−15 NF=6.40\times10^{-15}\,\text{N}; a=7.03×1015 m s−2a=7.03\times10^{15}\,\text{m s}^{-2}.

    Method: Convert V=1.80×103 VV=1.80\times10^3\,\text{V} and d=45.0×10−3 md=45.0\times10^{-3}\,\text{m}. Then E=V/d=4.00×104 V m−1E=V/d=4.00\times10^4\,\text{V m}^{-1}. The force magnitude is F=eE=(1.60×10−19)(4.00×104)=6.40×10−15 NF=eE=(1.60\times10^{-19})(4.00\times10^4)=6.40\times10^{-15}\,\text{N}. Using me=9.11×10−31 kgm_e=9.11\times10^{-31}\,\text{kg}, a=F/me=7.03×1015 m s−2a=F/m_e=7.03\times10^{15}\,\text{m s}^{-2}; its direction is opposite to the field.

Tier 3 · Hard

  1. 1. A proton enters at 3.00×106 m s−13.00\times10^6\,\text{m s}^{-1} perpendicular to a uniform electric field of strength 2.50×104 V m−12.50\times10^4\,\text{V m}^{-1}. The field region is 80.0 mm80.0\,\text{mm} long in the initial direction of travel. Determine the proton's deflection and the angle of its velocity to its original direction as it leaves. Use mp=1.67×10−27 kgm_p=1.67\times10^{-27}\,\text{kg}.[6 marks]

    Answer

    • Deflection 0.852 mm0.852\,\text{mm}; angle 1.22∘1.22^\circ.

    Method: The force is F=eEF=eE, so the transverse acceleration is a=eE/mp=(1.60×10−19)(2.50×104)/(1.67×10−27)=2.40×1012 m s−2a=eE/m_p=(1.60\times10^{-19})(2.50\times10^4)/(1.67\times10^{-27})=2.40\times10^{12}\,\text{m s}^{-2}. The transit time is fixed by horizontal motion: t=L/vx=0.0800/(3.00×106)=2.67×10−8 st=L/v_x=0.0800/(3.00\times10^6)=2.67\times10^{-8}\,\text{s}. Thus y=12at2=8.52×10−4 m=0.852 mmy=\tfrac12at^2=8.52\times10^{-4}\,\text{m}=0.852\,\text{mm}. The transverse exit speed is vy=at=6.39×104 m s−1v_y=at=6.39\times10^4\,\text{m s}^{-1}, so θ=tan⁡−1(vy/vx)=1.22∘\theta=\tan^{-1}(v_y/v_x)=1.22^\circ.

3.7.3.3 · Electric potential

  • Electric potential is work done per unit positive charge in bringing a small test charge from infinity; the zero of absolute potential is at infinity.
  • For a point charge, V=Q/(4πε0r)V=Q/(4\pi\varepsilon_0r). Potential is a scalar and carries the sign of the source charge.
  • The external work done in a slow transfer is ΔW=qΔV=q(Vfinal−Vinitial)\Delta W=q\Delta V=q(V_{\text{final}}-V_{\text{initial}}); the work done by the field is its negative.
  • No work is done along an equipotential. The field points towards decreasing potential, with E=ΔV/ΔrE=\Delta V/\Delta r in magnitude and potential difference given by area under an EE-rr graph.
  • A common error is to add electric-field magnitudes as scalars or to assume that V=0V=0 implies E=0E=0.

Tier 1 · Easy

  1. 1. Calculate the electric potential 0.200 m0.200\,\text{m} from an isolated point charge of +4.00 nC+4.00\,\text{nC}.[2 marks]

    Answer

    • +180 V+180\,\text{V}

    Method: Use V=kQ/rV=kQ/r. Therefore V=(8.99×109)(4.00×10−9)/0.200=179.8 VV=(8.99\times10^9)(4.00\times10^{-9})/0.200=179.8\,\text{V}. The source charge is positive, so V=+180 VV=+180\,\text{V} to three significant figures.

Tier 2 · Standard

  1. 1. A charge of −3.0 nC-3.0\,\text{nC} is moved slowly from a point at +120 V+120\,\text{V} to a point at −80 V-80\,\text{V}. Calculate the work done by the external force and state whether the charge's electric potential energy increases or decreases.[4 marks]

    Answer

    • External work +6.0×10−7 J+6.0\times10^{-7}\,\text{J}; potential energy increases.

    Method: The potential change is ΔV=Vf−Vi=−80−120=−200 V\Delta V=V_f-V_i=-80-120=-200\,\text{V}. For a slow transfer, external work equals the change in potential energy: ΔW=qΔV=(−3.0×10−9)(−200)=+6.0×10−7 J\Delta W=q\Delta V=(-3.0\times10^{-9})(-200)=+6.0\times10^{-7}\,\text{J}. The positive sign means the potential energy increases.

Tier 3 · Hard

  1. 1. Charges +8.0 nC+8.0\,\text{nC} and −2.0 nC-2.0\,\text{nC} are fixed 0.600 m0.600\,\text{m} apart. Find the point between them where the electric potential is zero. Determine the electric field strength there and the external work needed to bring a +3.0 nC+3.0\,\text{nC} charge slowly from infinity to that point.[6 marks]

    Answer

    • The point is 0.480 m0.480\,\text{m} from the positive charge; E=1.56×103 N C−1E=1.56\times10^3\,\text{N C}^{-1} towards the negative charge; external work 0 J0\,\text{J}.

    Method: Let the point be distance xx from the positive charge, so it is 0.600−x0.600-x from the negative charge. Set the scalar potential to zero: k[8.0×10−9/x−2.0×10−9/(0.600−x)]=0k[8.0\times10^{-9}/x-2.0\times10^{-9}/(0.600-x)]=0. Hence 8(0.600−x)=2x8(0.600-x)=2x, giving x=0.480 mx=0.480\,\text{m}. At this point both field vectors point from the positive charge towards the negative charge, so they add: E=k[8.0×10−9/(0.480)2+2.0×10−9/(0.120)2]=1.56×103 N C−1E=k[8.0\times10^{-9}/(0.480)^2+2.0\times10^{-9}/(0.120)^2]=1.56\times10^3\,\text{N C}^{-1}. Since V∞=Vpoint=0V_{\infty}=V_{\text{point}}=0, ΔW=qΔV=0\Delta W=q\Delta V=0 even though the field is not zero.

3.7.4.1 · Capacitance

  • Capacitance is charge stored per unit potential difference: C=Q/VC=Q/V. One farad is one coulomb per volt.
  • For a fixed capacitor, a graph of QQ against VV is a straight line through the origin with gradient CC.
  • At fixed capacitance, increasing potential difference increases the equal and opposite charge magnitudes on the plates proportionally.
  • Convert units before substitution: 1 μF=10−6 F1\,\mu\text{F}=10^{-6}\,\text{F}. A common error is to confuse charge on one plate with the net charge of the whole capacitor.

Tier 1 · Easy

  1. 1. A capacitor stores charge of 3.6 mC3.6\,\text{mC} at a potential difference of 12 V12\,\text{V}. Calculate its capacitance.[2 marks]

    Answer

    • 300 μF300\,\mu\text{F}

    Method: Use C=Q/VC=Q/V. Convert Q=3.6×10−3 CQ=3.6\times10^{-3}\,\text{C}, so C=(3.6×10−3)/12=3.0×10−4 F=300 μFC=(3.6\times10^{-3})/12=3.0\times10^{-4}\,\text{F}=300\,\mu\text{F}.

Tier 2 · Standard

  1. 1. A charge-potential-difference graph for a capacitor passes through the point (16.0 V,4.80 mC)(16.0\,\text{V},4.80\,\text{mC}). Determine its capacitance and the charge stored at 27.0 V27.0\,\text{V}.[3 marks]

    Answer

    • Capacitance 300 μF300\,\mu\text{F}; charge 8.10 mC8.10\,\text{mC}.

    Method: The gradient of a QQ-VV graph is CC. Therefore C=(4.80×10−3)/16.0=3.00×10−4 F=300 μFC=(4.80\times10^{-3})/16.0=3.00\times10^{-4}\,\text{F}=300\,\mu\text{F}. At 27.0 V27.0\,\text{V}, Q=CV=(3.00×10−4)(27.0)=8.10×10−3 C=8.10 mCQ=CV=(3.00\times10^{-4})(27.0)=8.10\times10^{-3}\,\text{C}=8.10\,\text{mC}.

Tier 3 · Hard

  1. 1. A 470 μF470\,\mu\text{F} capacitor is initially at 9.0 V9.0\,\text{V}. A charge-transfer process then moves 1.2×10161.2\times10^{16} electrons from its negatively charged plate to its positively charged plate while capacitance remains constant. Determine the new charge magnitude on each plate and the new potential difference.[5 marks]

    Answer

    • Charge 2.31 mC2.31\,\text{mC}; potential difference 4.91 V4.91\,\text{V}.

    Method: Initially Qi=CV=(470×10−6)(9.0)=4.23×10−3 CQ_i=CV=(470\times10^{-6})(9.0)=4.23\times10^{-3}\,\text{C}. The removed electron charge magnitude is ΔQ=ne=(1.2×1016)(1.60×10−19)=1.92×10−3 C\Delta Q=ne=(1.2\times10^{16})(1.60\times10^{-19})=1.92\times10^{-3}\,\text{C}. Hence Qf=Qi−ΔQ=2.31×10−3 CQ_f=Q_i-\Delta Q=2.31\times10^{-3}\,\text{C}. Since CC is unchanged, Vf=Qf/C=(2.31×10−3)/(470×10−6)=4.91 VV_f=Q_f/C=(2.31\times10^{-3})/(470\times10^{-6})=4.91\,\text{V}.

3.7.4.2 · Parallel plate capacitor

  • For parallel plates, C=ε0εrA/dC=\varepsilon_0\varepsilon_rA/d: capacitance increases with plate overlap area and relative permittivity, and decreases with separation.
  • A dielectric becomes polarised in the field. Bound charges create an opposing field, reducing the resultant field and potential difference for a given free charge.
  • If the capacitor remains connected to a fixed-voltage supply, adding a dielectric increases CC and therefore draws additional charge because Q=CVQ=CV.
  • Use the overlap area and separation in SI units. A common error is to say the dielectric increases charge when the isolated capacitor's free charge is actually fixed.

Tier 1 · Easy

  1. 1. Two parallel plates in air have overlap area 2.50×10−2 m22.50\times10^{-2}\,\text{m}^2 and separation 1.20 mm1.20\,\text{mm}. Calculate their capacitance.[2 marks]

    Answer

    • 1.84×10−10 F1.84\times10^{-10}\,\text{F}

    Method: For air take εr=1\varepsilon_r=1. Convert d=1.20×10−3 md=1.20\times10^{-3}\,\text{m} and use C=ε0A/dC=\varepsilon_0A/d. Thus C=(8.85×10−12)(2.50×10−2)/(1.20×10−3)=1.84×10−10 FC=(8.85\times10^{-12})(2.50\times10^{-2})/(1.20\times10^{-3})=1.84\times10^{-10}\,\text{F}.

Tier 2 · Standard

  1. 1. A parallel-plate capacitor has area 1.80×10−2 m21.80\times10^{-2}\,\text{m}^2, separation 0.800 mm0.800\,\text{mm} and is connected to a 120 V120\,\text{V} supply. A dielectric of relative permittivity 3.403.40 completely fills the gap. Determine the new capacitance and charge stored.[4 marks]

    Answer

    • Capacitance 6.77×10−10 F6.77\times10^{-10}\,\text{F}; charge 8.12×10−8 C8.12\times10^{-8}\,\text{C}.

    Method: Convert d=8.00×10−4 md=8.00\times10^{-4}\,\text{m}. With the dielectric, C=ε0εrA/d=(8.85×10−12)(3.40)(1.80×10−2)/(8.00×10−4)=6.77×10−10 FC=\varepsilon_0\varepsilon_rA/d=(8.85\times10^{-12})(3.40)(1.80\times10^{-2})/(8.00\times10^{-4})=6.77\times10^{-10}\,\text{F}. The supply keeps VV fixed, so Q=CV=(6.77×10−10)(120)=8.12×10−8 CQ=CV=(6.77\times10^{-10})(120)=8.12\times10^{-8}\,\text{C}.

Tier 3 · Hard

  1. 1. A designer needs a 2.00 nF2.00\,\text{nF} parallel-plate capacitor using a dielectric with relative permittivity 4.504.50 and thickness 0.750 mm0.750\,\text{mm}. Calculate the required overlap area. Explain at the molecular level why the dielectric increases capacitance.[6 marks]

    Answer

    • Required area 3.77×10−2 m23.77\times10^{-2}\,\text{m}^2; polarisation reduces the field and potential difference for a given free charge.

    Method: Rearrange C=ε0εrA/dC=\varepsilon_0\varepsilon_rA/d to A=Cd/(ε0εr)A=Cd/(\varepsilon_0\varepsilon_r). Hence A=(2.00×10−9)(0.750×10−3)/[(8.85×10−12)(4.50)]=3.77×10−2 m2A=(2.00\times10^{-9})(0.750\times10^{-3})/[(8.85\times10^{-12})(4.50)]=3.77\times10^{-2}\,\text{m}^2. In the dielectric, permanent polar molecules rotate, or induced dipoles align, so bound surface charges form. Their field opposes the field due to the free plate charges. For a fixed free charge this lowers VV, and because C=Q/VC=Q/V, the capacitance is larger.

3.7.4.3 · Energy stored by a capacitor

  • The energy stored is the work done separating charge: E=12QV=12CV2=12Q2/CE=\tfrac12QV=\tfrac12CV^2=\tfrac12Q^2/C.
  • On a graph of potential difference against charge, stored energy is the area under the line; for a linear capacitor this is a triangle.
  • Choose the energy form that matches the quantities given and retain the square on VV or QQ where required.
  • When voltage falls, energy released is the difference between initial and final stored energies, not simply QΔVQ\Delta V. This is a common marking error.

Tier 1 · Easy

  1. 1. A capacitor stores 2.0 mC2.0\,\text{mC} at a potential difference of 9.0 V9.0\,\text{V}. Calculate its stored energy.[2 marks]

    Answer

    • 9.0 mJ9.0\,\text{mJ}

    Method: Use E=12QVE=\tfrac12QV. Thus E=12(2.0×10−3)(9.0)=9.0×10−3 J=9.0 mJE=\tfrac12(2.0\times10^{-3})(9.0)=9.0\times10^{-3}\,\text{J}=9.0\,\text{mJ}.

Tier 2 · Standard

  1. 1. The potential difference across a 150 μF150\,\mu\text{F} capacitor falls from 20.0 V20.0\,\text{V} to 8.0 V8.0\,\text{V}. Determine the energy transferred from the capacitor.[3 marks]

    Answer

    • 2.52×10−2 J2.52\times10^{-2}\,\text{J}

    Method: Energy transferred is the decrease in stored energy: ΔE=12C(Vi2−Vf2)\Delta E=\tfrac12C(V_i^2-V_f^2). Therefore ΔE=12(150×10−6)[(20.0)2−(8.0)2]=2.52×10−2 J\Delta E=\tfrac12(150\times10^{-6})[(20.0)^2-(8.0)^2]=2.52\times10^{-2}\,\text{J}.

Tier 3 · Hard

  1. 1. A 330 μF330\,\mu\text{F} capacitor is charged from zero to 24.0 V24.0\,\text{V} through a resistor by an ideal constant-voltage supply. Calculate the work done by the supply, the final stored energy, and the mean power dissipated in the resistor if charging takes 0.800 s0.800\,\text{s}.[5 marks]

    Answer

    • Supply work 0.190 J0.190\,\text{J}; stored energy 0.0950 J0.0950\,\text{J}; mean resistor power 0.119 W0.119\,\text{W}.

    Method: The final charge is Q=CV=(330×10−6)(24.0)=7.92×10−3 CQ=CV=(330\times10^{-6})(24.0)=7.92\times10^{-3}\,\text{C}. The supply works at fixed voltage, so Wsupply=QV=CV2=(330×10−6)(24.0)2=0.19008 JW_{\text{supply}}=QV=CV^2=(330\times10^{-6})(24.0)^2=0.19008\,\text{J}. The capacitor stores E=12CV2=0.09504 JE=\tfrac12CV^2=0.09504\,\text{J}, equal to the triangular area under its VV-QQ graph. The remainder 0.19008−0.09504=0.09504 J0.19008-0.09504=0.09504\,\text{J} is dissipated, so mean power is 0.09504/0.800=0.1188 W0.09504/0.800=0.1188\,\text{W}.

3.7.4.4 · Capacitor charge and discharge

  • The time constant is τ=RC\tau=RC. During discharge, Q=Q0e−t/RCQ=Q_0e^{-t/RC}, with corresponding exponential equations for VV and the magnitude of II.
  • After one time constant a discharging quantity is e−1=0.368e^{-1}=0.368 of its initial value; the half-life is T1/2=0.69RCT_{1/2}=0.69RC.
  • During charging, Q=Q0(1−e−t/RC)Q=Q_0(1-e^{-t/RC}); charge and voltage rise towards a limit while current falls towards zero.
  • For discharge, plotting ln⁡Q\ln Q against tt gives a straight line of gradient −1/RC-1/RC. A common error is to treat one time constant as the time for complete discharge.
  • On graph questions, a tangent gives instantaneous current through I=dQ/dtI=\mathrm{d}Q/\mathrm{d}t, while area under an II-tt graph gives transferred charge.

Tier 1 · Easy

  1. 1. A 220 μF220\,\mu\text{F} capacitor discharges through a 47 kΩ47\,\text{k}\Omega resistor. Calculate the time constant and state the fraction of initial charge remaining after this time.[2 marks]

    Answer

    • Time constant 10.3 s10.3\,\text{s}; fraction remaining 0.3680.368.

    Method: Convert R=47×103 ΩR=47\times10^3\,\Omega and C=220×10−6 FC=220\times10^{-6}\,\text{F}. Then τ=RC=(47×103)(220×10−6)=10.34 s\tau=RC=(47\times10^3)(220\times10^{-6})=10.34\,\text{s}. At t=τt=\tau, Q/Q0=e−1=0.368Q/Q_0=e^{-1}=0.368.

Tier 2 · Standard

  1. 1. A capacitor initially stores 6.00 mC6.00\,\text{mC} and discharges through an 82.0 kΩ82.0\,\text{k}\Omega resistor. Its capacitance is 100 μF100\,\mu\text{F}. Determine the charge remaining after 12.0 s12.0\,\text{s}.[3 marks]

    Answer

    • 1.39 mC1.39\,\text{mC}

    Method: The time constant is RC=(82.0×103)(100×10−6)=8.20 sRC=(82.0\times10^3)(100\times10^{-6})=8.20\,\text{s}. Use Q=Q0e−t/RCQ=Q_0e^{-t/RC}: Q=(6.00×10−3)e−12.0/8.20=1.3887×10−3 C=1.39 mCQ=(6.00\times10^{-3})e^{-12.0/8.20}=1.3887\times10^{-3}\,\text{C}=1.39\,\text{mC}.

Tier 3 · Hard

  1. 1. During discharge, charge falls from 8.0 mC8.0\,\text{mC} to 1.6 mC1.6\,\text{mC} in 14.0 s14.0\,\text{s}. The capacitance is 120 μF120\,\mu\text{F}. Determine the time constant, the resistance, and the charge after one further time constant. State the gradient of a graph of ln⁡Q\ln Q against tt.[6 marks]

    Answer

    • τ=8.70 s\tau=8.70\,\text{s}; R=72.5 kΩR=72.5\,\text{k}\Omega; later charge 0.589 mC0.589\,\text{mC}; gradient −0.115 s−1-0.115\,\text{s}^{-1}.

    Method: Use Q/Q0=e−t/τQ/Q_0=e^{-t/\tau}. Here 1.6/8.0=0.2001.6/8.0=0.200, so ln⁡(0.200)=−14.0/τ\ln(0.200)=-14.0/\tau and τ=14.0/ln⁡5=8.70 s\tau=14.0/\ln 5=8.70\,\text{s}. Since τ=RC\tau=RC, R=8.70/(120×10−6)=7.25×104 Ω=72.5 kΩR=8.70/(120\times10^{-6})=7.25\times10^4\,\Omega=72.5\,\text{k}\Omega. One further time constant multiplies charge by e−1e^{-1}, giving 1.6/e=0.589 mC1.6/e=0.589\,\text{mC}. Taking logs gives ln⁡Q=ln⁡Q0−t/RC\ln Q=\ln Q_0-t/RC, so the gradient is −1/τ=−0.115 s−1-1/\tau=-0.115\,\text{s}^{-1}.

3.7.5.1 · Magnetic flux density

  • For a straight wire perpendicular to a magnetic field, F=BIlF=BIl. Magnetic flux density is therefore B=F/(Il)B=F/(Il).
  • One tesla is the flux density that gives a force of 1 N1\,\text{N} on a 1 m1\,\text{m} wire carrying 1 A1\,\text{A} perpendicular to the field.
  • Fleming's left-hand rule gives force, field and conventional-current directions; electron flow is opposite to conventional current.
  • In a top-pan-balance experiment, convert the change in mass reading to force using F=ΔmgF=\Delta mg. A common error is to substitute grams directly into the force equation.

Tier 1 · Easy

  1. 1. A 0.080 m0.080\,\text{m} wire carries 3.2 A3.2\,\text{A} perpendicular to a magnetic field of flux density 0.45 T0.45\,\text{T}. Calculate the force on the wire.[2 marks]

    Answer

    • 0.12 N0.12\,\text{N}

    Method: Use F=BIl=(0.45)(3.2)(0.080)=0.1152 NF=BIl=(0.45)(3.2)(0.080)=0.1152\,\text{N}. To two significant figures the force is 0.12 N0.12\,\text{N}.

Tier 2 · Standard

  1. 1. A horizontal wire of length 0.120 m0.120\,\text{m} carries 4.00 A4.00\,\text{A} perpendicular to a magnetic field. Switching on the current changes the balance reading by 6.50 g6.50\,\text{g}. Determine the magnetic flux density.[3 marks]

    Answer

    • 0.133 T0.133\,\text{T}

    Method: The stated change from the zero-current reading corresponds to magnetic force F=Δmg=(6.50×10−3)(9.81)=0.0638 NF=\Delta mg=(6.50\times10^{-3})(9.81)=0.0638\,\text{N}. Then B=F/(Il)=0.0638/[(4.00)(0.120)]=0.133 TB=F/(Il)=0.0638/[(4.00)(0.120)]=0.133\,\text{T}.

Tier 3 · Hard

  1. 1. In a balance experiment, a perpendicular wire of active length 0.0850 m0.0850\,\text{m} gives a straight-line graph of balance-reading change against current with gradient 1.75 g A−11.75\,\text{g A}^{-1}. Determine the flux density and predict the reading change at 5.20 A5.20\,\text{A}. Explain what a non-zero vertical intercept would suggest.[5 marks]

    Answer

    • B=0.202 TB=0.202\,\text{T}; reading change 9.10 g9.10\,\text{g}; a non-zero intercept indicates a zero offset or other systematic force.

    Method: Convert the gradient to force per current: (1.75×10−3)(9.81)=1.71675×10−2 N A−1(1.75\times10^{-3})(9.81)=1.71675\times10^{-2}\,\text{N A}^{-1}. Since F/I=BlF/I=Bl, B=(1.71675×10−2)/0.0850=0.202 TB=(1.71675\times10^{-2})/0.0850=0.202\,\text{T}. At 5.20 A5.20\,\text{A} the mass-reading change is (1.75)(5.20)=9.10 g(1.75)(5.20)=9.10\,\text{g}. The magnetic relationship predicts zero force at zero current, so a non-zero intercept indicates a systematic zero error or a current-independent force, not a change in BB.

3.7.5.2 · Moving charges in a magnetic field

  • A charge moving perpendicular to a magnetic field experiences force F=BQvF=BQv. The force is perpendicular to both velocity and field.
  • Use Fleming's left-hand rule for positive conventional current; reverse the predicted force for a negative particle.
  • Because magnetic force is perpendicular to velocity, it does no work and changes direction without changing speed or kinetic energy.
  • For circular motion, BQv=mv2/rBQv=mv^2/r, so r=mv/(BQ)r=mv/(BQ). A common error is to include weight or use the particle's diameter as rr.
  • In a cyclotron, the non-relativistic orbital frequency is f=BQ/(2πm)f=BQ/(2\pi m) and is independent of radius and speed.

Tier 1 · Easy

  1. 1. A proton moves at 4.0×106 m s−14.0\times10^6\,\text{m s}^{-1} perpendicular to a 0.25 T0.25\,\text{T} magnetic field. Calculate the magnetic force magnitude.[2 marks]

    Answer

    • 1.6×10−13 N1.6\times10^{-13}\,\text{N}

    Method: Use F=BQv=(0.25)(1.60×10−19)(4.0×106)=1.6×10−13 NF=BQv=(0.25)(1.60\times10^{-19})(4.0\times10^6)=1.6\times10^{-13}\,\text{N}.

Tier 2 · Standard

  1. 1. A proton of speed 3.20×106 m s−13.20\times10^6\,\text{m s}^{-1} enters a uniform 0.480 T0.480\,\text{T} magnetic field perpendicular to the field. Determine the radius of its path. Use mp=1.67×10−27 kgm_p=1.67\times10^{-27}\,\text{kg}.[3 marks]

    Answer

    • 6.96×10−2 m6.96\times10^{-2}\,\text{m}

    Method: Magnetic force supplies centripetal force: BQv=mv2/rBQv=mv^2/r. Hence r=mv/(BQ)=(1.67×10−27)(3.20×106)/[(0.480)(1.60×10−19)]=6.96×10−2 mr=mv/(BQ)=(1.67\times10^{-27})(3.20\times10^6)/[(0.480)(1.60\times10^{-19})]=6.96\times10^{-2}\,\text{m}.

Tier 3 · Hard

  1. 1. An alpha particle of mass 6.64×10−27 kg6.64\times10^{-27}\,\text{kg} and charge +3.20×10−19 C+3.20\times10^{-19}\,\text{C} reaches radius 0.450 m0.450\,\text{m} in a cyclotron with magnetic flux density 0.800 T0.800\,\text{T}. Determine its speed, kinetic energy in MeV\text{MeV}, and cyclotron frequency.[6 marks]

    Answer

    • Speed 1.73×107 m s−11.73\times10^7\,\text{m s}^{-1}; kinetic energy 6.25 MeV6.25\,\text{MeV}; frequency 6.14 MHz6.14\,\text{MHz}.

    Method: For circular motion, BQv=mv2/rBQv=mv^2/r, so v=BQr/m=(0.800)(3.20×10−19)(0.450)/(6.64×10−27)=1.73×107 m s−1v=BQr/m=(0.800)(3.20\times10^{-19})(0.450)/(6.64\times10^{-27})=1.73\times10^7\,\text{m s}^{-1}. Then Ek=12mv2=9.99×10−13 JE_k=\tfrac12mv^2=9.99\times10^{-13}\,\text{J}. Since 1 MeV=1.60×10−13 J1\,\text{MeV}=1.60\times10^{-13}\,\text{J}, this is 6.25 MeV6.25\,\text{MeV}. The period follows from 2πr/v2\pi r/v, giving f=BQ/(2πm)=(0.800)(3.20×10−19)/[2π(6.64×10−27)]=6.14×106 Hzf=BQ/(2\pi m)=(0.800)(3.20\times10^{-19})/[2\pi(6.64\times10^{-27})]=6.14\times10^6\,\text{Hz}.

3.7.5.3 · Magnetic flux and flux linkage

  • Magnetic flux is Φ=BA\Phi=BA when the field is normal to area AA, and is measured in webers.
  • For a flat coil at angle θ\theta between the field and the coil's normal, flux linkage is NΦ=BANcos⁡θN\Phi=BAN\cos\theta.
  • Flux linkage counts the flux through every turn; changing NN, BB, AA or orientation changes it.
  • A negative value of NΦN\Phi indicates that flux passes through the coil in the direction opposite to the chosen positive normal. A common error is to use the angle to the plane instead of the normal.

Tier 1 · Easy

  1. 1. Magnetic flux passes normally through a flat surface of area 1.5×10−2 m21.5\times10^{-2}\,\text{m}^2. The flux density is 0.32 T0.32\,\text{T}. Calculate the flux.[2 marks]

    Answer

    • 4.8×10−3 Wb4.8\times10^{-3}\,\text{Wb}

    Method: With the field normal to the area, Φ=BA=(0.32)(1.5×10−2)=4.8×10−3 Wb\Phi=BA=(0.32)(1.5\times10^{-2})=4.8\times10^{-3}\,\text{Wb}.

Tier 2 · Standard

  1. 1. A 240240-turn coil of area 3.50×10−3 m23.50\times10^{-3}\,\text{m}^2 is in a 0.180 T0.180\,\text{T} field. The field makes an angle of 35.0∘35.0^\circ with the normal to the coil. Determine the flux linkage.[3 marks]

    Answer

    • 0.124 Wb turns0.124\,\text{Wb turns}

    Method: Use NΦ=BANcos⁡θN\Phi=BAN\cos\theta. Thus NΦ=(0.180)(3.50×10−3)(240)cos⁡35.0∘=0.1239 Wb turnsN\Phi=(0.180)(3.50\times10^{-3})(240)\cos35.0^\circ=0.1239\,\text{Wb turns}, which is 0.124 Wb turns0.124\,\text{Wb turns}.

Tier 3 · Hard

  1. 1. A 500500-turn search coil of area 8.0×10−4 m28.0\times10^{-4}\,\text{m}^2 is in a uniform 0.120 T0.120\,\text{T} field. Its chosen normal is rotated from 20.0∘20.0^\circ to 110∘110^\circ relative to the field. Calculate the initial and final flux linkages and their signed change.[5 marks]

    Answer

    • Initial +4.51×10−2 Wb turns+4.51\times10^{-2}\,\text{Wb turns}; final −1.64×10−2 Wb turns-1.64\times10^{-2}\,\text{Wb turns}; change −6.15×10−2 Wb turns-6.15\times10^{-2}\,\text{Wb turns}.

    Method: Use NΦ=BANcos⁡θN\Phi=BAN\cos\theta with BAN=(0.120)(8.0×10−4)(500)=0.0480 Wb turnsBAN=(0.120)(8.0\times10^{-4})(500)=0.0480\,\text{Wb turns}. Initially, NΦi=0.0480cos⁡20.0∘=+4.51×10−2 Wb turnsN\Phi_i=0.0480\cos20.0^\circ=+4.51\times10^{-2}\,\text{Wb turns}. Finally, NΦf=0.0480cos⁡110∘=−1.64×10−2 Wb turnsN\Phi_f=0.0480\cos110^\circ=-1.64\times10^{-2}\,\text{Wb turns}. Therefore Δ(NΦ)=NΦf−NΦi=−6.15×10−2 Wb turns\Delta(N\Phi)=N\Phi_f-N\Phi_i=-6.15\times10^{-2}\,\text{Wb turns}; the negative sign records reversal relative to the chosen normal.

3.7.5.4 · Electromagnetic induction

  • Faraday's law states that the induced emf magnitude equals the rate of change of flux linkage: ∣ε∣=∣Δ(NΦ)/Δt∣|\varepsilon|=|\Delta(N\Phi)/\Delta t|.
  • Lenz's law gives the direction: the induced current produces effects that oppose the change in flux that caused it.
  • A straight conductor cutting field lines has an induced emf; a stationary search coil in a steady field has none unless its flux linkage changes.
  • For a coil rotating uniformly, ε=BANωsin⁡ωt\varepsilon=BAN\omega\sin\omega t and the peak emf is BANωBAN\omega.
  • A common error is to say induced current opposes the field; it opposes the change in flux, and a current exists only when the circuit is complete.

Tier 1 · Easy

  1. 1. State Faraday's law and Lenz's law for electromagnetic induction.[2 marks]

    Answer

    • Faraday's law: induced emf equals the rate of change of flux linkage. Lenz's law: the induced current acts to oppose the change in flux producing it.

    Method: For Faraday's law, state the rate of change of flux linkage, not merely flux. For Lenz's law, identify opposition to the change that causes the induction; this determines the polarity or current direction.

Tier 2 · Standard

  1. 1. The magnetic flux through each turn of a 250250-turn coil decreases uniformly from 3.2 mWb3.2\,\text{mWb} to 0.80 mWb0.80\,\text{mWb} in 40 ms40\,\text{ms}. Determine the induced emf magnitude and state how Lenz's law fixes its polarity.[3 marks]

    Answer

    • 15 V15\,\text{V}; the polarity drives a current whose magnetic field opposes the decrease in flux.

    Method: Faraday's law gives ∣ε∣=N∣ΔΦ∣/Δt=250(3.2−0.80)×10−3/(40×10−3)=15 V|\varepsilon|=N|\Delta\Phi|/\Delta t=250(3.2-0.80)\times10^{-3}/(40\times10^{-3})=15\,\text{V}. By Lenz's law, the induced polarity is such that any induced current produces flux in the original direction, opposing the stated decrease.

Tier 3 · Hard

  1. 1. A 180180-turn coil of area 6.50×10−3 m26.50\times10^{-3}\,\text{m}^2 rotates at 50.0 Hz50.0\,\text{Hz} in a uniform 0.240 T0.240\,\text{T} field. At t=0t=0 its flux linkage is maximum and positive. Calculate the peak emf and the emf magnitude at t=2.50 mst=2.50\,\text{ms}. Explain the origin and direction of the emf using Faraday's and Lenz's laws.[5 marks]

    Answer

    • Peak emf 88.2 V88.2\,\text{V}; emf magnitude at 2.50 ms2.50\,\text{ms} is 62.4 V62.4\,\text{V}.

    Method: The angular speed is ω=2πf=314 rad s−1\omega=2\pi f=314\,\text{rad s}^{-1}. The flux linkage is BANcos⁡ωtBAN\cos\omega t, so Faraday's law gives ε=BANωsin⁡ωt\varepsilon=BAN\omega\sin\omega t in magnitude. Thus ε0=(0.240)(6.50×10−3)(180)(314)=88.2 V\varepsilon_0=(0.240)(6.50\times10^{-3})(180)(314)=88.2\,\text{V}. At t=2.50×10−3 st=2.50\times10^{-3}\,\text{s}, ωt=0.785 rad\omega t=0.785\,\text{rad} and ∣ε∣=88.2sin⁡0.785=62.4 V|\varepsilon|=88.2\sin0.785=62.4\,\text{V}. The emf exists because rotation changes flux linkage; its polarity, from Lenz's law, drives a current whose magnetic effect opposes that change.

3.7.5.5 · Alternating currents

  • For a sinusoidal waveform, Irms=I0/2I_{\mathrm{rms}}=I_0/\sqrt2 and Vrms=V0/2V_{\mathrm{rms}}=V_0/\sqrt2, where subscript 00 denotes peak value.
  • Peak-to-peak value is twice the peak value. Mains voltage is quoted as an rms value, not a peak value.
  • An rms current has the same mean heating effect in a resistor as a direct current of that value; for a resistive load, mean power is VrmsIrmsV_{\mathrm{rms}}I_{\mathrm{rms}}.
  • On an oscilloscope, amplitude equals vertical divisions times volts per division; period equals horizontal divisions times time per division, and f=1/Tf=1/T.
  • A common error is to apply the 2\sqrt2 relationships to non-sinusoidal waveforms or to confuse peak with peak-to-peak.

Tier 1 · Easy

  1. 1. A sinusoidal voltage has peak value 12.0 V12.0\,\text{V}. Calculate its rms value.[1 mark]

    Answer

    • 8.49 V8.49\,\text{V}

    Method: For a sinusoid, Vrms=V0/2=12.0/2=8.49 VV_{\mathrm{rms}}=V_0/\sqrt2=12.0/\sqrt2=8.49\,\text{V}.

Tier 2 · Standard

  1. 1. A sinusoidal trace spans 6.46.4 horizontal divisions per cycle at 0.50 ms div−10.50\,\text{ms div}^{-1}. Its peak-to-peak height is 5.65.6 vertical divisions at 2.0 V div−12.0\,\text{V div}^{-1}. Determine the frequency, peak voltage and rms voltage.[4 marks]

    Answer

    • Frequency 3.1×102 Hz3.1\times10^2\,\text{Hz}; peak voltage 5.6 V5.6\,\text{V}; rms voltage 4.0 V4.0\,\text{V}.

    Method: The period is T=(6.4)(0.50 ms)=3.2×10−3 sT=(6.4)(0.50\,\text{ms})=3.2\times10^{-3}\,\text{s}, so f=1/T=3.125×102 Hzf=1/T=3.125\times10^2\,\text{Hz}. The peak-to-peak voltage is (5.6)(2.0)=11.2 V(5.6)(2.0)=11.2\,\text{V}, hence V0=11.2/2=5.6 VV_0=11.2/2=5.6\,\text{V}. For the sinusoid, Vrms=5.6/2=3.96 VV_{\mathrm{rms}}=5.6/\sqrt2=3.96\,\text{V}. Values to two significant figures are 3.1×102 Hz3.1\times10^2\,\text{Hz} and 4.0 V4.0\,\text{V}.

Tier 3 · Hard

  1. 1. A resistive heater rated at 1.80 kW1.80\,\text{kW} operates from a sinusoidal 230 V rms230\,\text{V rms} supply. Determine the peak and peak-to-peak supply voltages, the rms and peak currents, and the energy transferred in 12.0 min12.0\,\text{min}.[6 marks]

    Answer

    • V0=325 VV_0=325\,\text{V}; Vpp=651 VV_{\mathrm{pp}}=651\,\text{V}; Irms=7.83 AI_{\mathrm{rms}}=7.83\,\text{A}; I0=11.1 AI_0=11.1\,\text{A}; energy 1.30 MJ1.30\,\text{MJ}.

    Method: For a sinusoid, V0=2Vrms=2(230)=325 VV_0=\sqrt2V_{\mathrm{rms}}=\sqrt2(230)=325\,\text{V} and Vpp=2V0=651 VV_{\mathrm{pp}}=2V_0=651\,\text{V}. For a resistive heater, P=VrmsIrmsP=V_{\mathrm{rms}}I_{\mathrm{rms}}, so Irms=1800/230=7.83 AI_{\mathrm{rms}}=1800/230=7.83\,\text{A} and I0=2Irms=11.1 AI_0=\sqrt2I_{\mathrm{rms}}=11.1\,\text{A}. Convert 12.0 min=720 s12.0\,\text{min}=720\,\text{s}; then energy E=Pt=(1800)(720)=1.296×106 J=1.30 MJE=Pt=(1800)(720)=1.296\times10^6\,\text{J}=1.30\,\text{MJ}.

3.7.5.6 · The operation of a transformer

  • For an ideal transformer, Ns/Np=Vs/VpN_s/N_p=V_s/V_p. A step-up transformer has more secondary turns and a larger secondary voltage.
  • Efficiency is η=Pout/Pin=IsVs/(IpVp)\eta=P_{\mathrm{out}}/P_{\mathrm{in}}=I_sV_s/(I_pV_p); an ideal transformer conserves power, so stepping voltage up steps current down.
  • Energy losses include resistive heating in windings, eddy currents and hysteresis in the core, and flux leakage.
  • Laminating and electrically insulating the core restricts eddy-current loops; a suitable soft magnetic core reduces hysteresis loss.
  • High-voltage transmission reduces current for a given power and therefore reduces cable loss Ploss=I2RP_{\mathrm{loss}}=I^2R. A common error is to calculate loss using transmitted voltage across the cable resistance.

Tier 1 · Easy

  1. 1. An ideal transformer has 300300 primary turns and 12001200 secondary turns. The primary voltage is 24 V24\,\text{V}. Calculate the secondary voltage.[2 marks]

    Answer

    • 96 V96\,\text{V}

    Method: Use Vs/Vp=Ns/NpV_s/V_p=N_s/N_p. Hence Vs=24(1200/300)=96 VV_s=24(1200/300)=96\,\text{V}.

Tier 2 · Standard

  1. 1. A transformer takes 1.80 A1.80\,\text{A} from a 230 V230\,\text{V} supply and delivers 24.0 V24.0\,\text{V} at 84.0%84.0\% efficiency. Determine the secondary current and the power dissipated in the transformer.[4 marks]

    Answer

    • Secondary current 14.5 A14.5\,\text{A}; dissipated power 66.2 W66.2\,\text{W}.

    Method: Input power is Pin=IpVp=(1.80)(230)=414 WP_{\mathrm{in}}=I_pV_p=(1.80)(230)=414\,\text{W}. Output power is Pout=ηPin=(0.840)(414)=347.76 WP_{\mathrm{out}}=\eta P_{\mathrm{in}}=(0.840)(414)=347.76\,\text{W}. Thus Is=Pout/Vs=347.76/24.0=14.49 AI_s=P_{\mathrm{out}}/V_s=347.76/24.0=14.49\,\text{A}. The dissipated power is Pin−Pout=414−347.76=66.24 WP_{\mathrm{in}}-P_{\mathrm{out}}=414-347.76=66.24\,\text{W}.

Tier 3 · Hard

  1. 1. A station transmits 2.50 MW2.50\,\text{MW} through cables of total resistance 3.20 Ω3.20\,\Omega. Compare the cable power losses when transmission voltage is 25.0 kV25.0\,\text{kV} and 250 kV250\,\text{kV}. Explain the transformer's role and how core construction reduces eddy-current loss.[6 marks]

    Answer

    • Loss at 25.0 kV25.0\,\text{kV} is 32.0 kW32.0\,\text{kW}; loss at 250 kV250\,\text{kV} is 320 W320\,\text{W}, one hundredth as large. A step-up transformer reduces current, and insulated laminations restrict eddy currents.

    Method: Assuming the transmitted power is the stated 2.50 MW2.50\,\text{MW}, at 25.0 kV25.0\,\text{kV} the current is I=P/V=(2.50×106)/(25.0×103)=100 AI=P/V=(2.50\times10^6)/(25.0\times10^3)=100\,\text{A}, so Ploss=I2R=(100)2(3.20)=3.20×104 W=32.0 kWP_{\mathrm{loss}}=I^2R=(100)^2(3.20)=3.20\times10^4\,\text{W}=32.0\,\text{kW}. At 250 kV250\,\text{kV}, I=10.0 AI=10.0\,\text{A} and loss is (10.0)2(3.20)=320 W(10.0)^2(3.20)=320\,\text{W}, a factor of 100100 smaller. A step-up transformer uses Ns/Np=Vs/VpN_s/N_p=V_s/V_p to raise voltage and, approximately conserving power, reduce current. Alternating flux can induce circulating eddy currents in a solid core; thin mutually insulated laminations break the current paths, raising their resistance and reducing I2RI^2R heating.