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3. A solenoid is used to generate a magnetic field. The solenoid has an inner radius a, length, and N total turns of wire. A power supply, not shown, is connected to the solenoid and generates current I, as shown in the figure on the left above. The x-axis runs along the axis of the solenoid. Point P is in the middle of the solenoid at the origin of the xyz-coordinate system, as shown in the cutaway view on the right above. Assume >> a. (a) Select the correct direction of the magnetic field at point P. +x-direction +z-direction -z-direction -x-direction Justify your selection. (b) +y-direction -y-direction i. On the cutaway view below, clearly draw an amperian loop that can be used to determine the magnetic field at point P at the center of the solenoid. Axis of Solenoid Cutaway View

3. A solenoid is used to generate a magnetic field. The solenoid has an inner radius a, length, and N total turns of wire. A power supply, not shown, is connected to the solenoid and generates current I, as shown in the figure on the left above. The x-axis runs along the axis of the solenoid. Point P is in the middle of the solenoid at the origin of the xyz-coordinate system, as shown in the cutaway view on the right above. Assume >> a. (a) Select the correct direction of the magnetic field at point P. +x-direction +z-direction -z-direction -x-direction Justify your selection. (b) +y-direction -y-direction i. On the cutaway view below, clearly draw an amperian loop that can be used to determine the magnetic field at point P at the center of the solenoid. Axis of Solenoid Cutaway View

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(b) Note: Figures not drawn to scale. 3. A solenoid is used to generate a magnetic field. The solenoid has an inner radius a, length, and N total turns of wire. A power supply, not shown, is connected to the solenoid and generates current I, as shown in the figure on the left above. The x-axis runs along the axis of the solenoid. Point P is in the middle of the solenoid at the origin of the xyz-coordinate system, as shown in the cutaway view on the right above. Assume >> a. (a) Select the correct direction of the magnetic field at point P. +x-direction +z-direction -z-direction -x-direction Justify your selection. ri gle Circle Kectung.c A on A = r² C = 2nr i. On the cutaway view below, clearly draw an amperian loop that can be used to determine the magnetic field at point P at the center of the solenoid. 1 =r6 Rectangular Solid Vlh Cy.inder Sphere 1 GEOMETRY AND TRIGONOMETRY Cutaway View ii. Use Ampere's law to derive an expression for the magnetic field strength at point P. Express your answer in terms of I, , N, a, and physical constants, as appropriate. V = xr²t SL. 2m. V=R² S = 4; ² Rig Triangle 2 + b² = c² a == sine = Solenoid cose= tane= +y-direction -y-direction b C Axis of Solenoid ADVANCED PLACEMENT PHYSICS C EQUATIONS A-area C=circumference V-volume S-sur. ce area sC M h-hei € = le L W = IUL r =1.u.u. S Axis of Solenoid arc leg = «fg.C 0 Cutaway View b 90% a CALCULUS af df df du du dx (x²)=x²-1 (²) = aeª (In cc) = [sin (ax)] = acos (ax) cs(x)]=-a.ax, 1 [x²x = 1 + 1² n+1 Se dx = - ex cbc √x + a +1 -1 = In|x + al [cos(x) dx = sin(x) a [sin (ax) dx = cos(x) VECTOR PRODUCTS A-B = AB cose Ax B= ABsine
Transcribed Image Text:(b) Note: Figures not drawn to scale. 3. A solenoid is used to generate a magnetic field. The solenoid has an inner radius a, length, and N total turns of wire. A power supply, not shown, is connected to the solenoid and generates current I, as shown in the figure on the left above. The x-axis runs along the axis of the solenoid. Point P is in the middle of the solenoid at the origin of the xyz-coordinate system, as shown in the cutaway view on the right above. Assume >> a. (a) Select the correct direction of the magnetic field at point P. +x-direction +z-direction -z-direction -x-direction Justify your selection. ri gle Circle Kectung.c A on A = r² C = 2nr i. On the cutaway view below, clearly draw an amperian loop that can be used to determine the magnetic field at point P at the center of the solenoid. 1 =r6 Rectangular Solid Vlh Cy.inder Sphere 1 GEOMETRY AND TRIGONOMETRY Cutaway View ii. Use Ampere's law to derive an expression for the magnetic field strength at point P. Express your answer in terms of I, , N, a, and physical constants, as appropriate. V = xr²t SL. 2m. V=R² S = 4; ² Rig Triangle 2 + b² = c² a == sine = Solenoid cose= tane= +y-direction -y-direction b C Axis of Solenoid ADVANCED PLACEMENT PHYSICS C EQUATIONS A-area C=circumference V-volume S-sur. ce area sC M h-hei € = le L W = IUL r =1.u.u. S Axis of Solenoid arc leg = «fg.C 0 Cutaway View b 90% a CALCULUS af df df du du dx (x²)=x²-1 (²) = aeª (In cc) = [sin (ax)] = acos (ax) cs(x)]=-a.ax, 1 [x²x = 1 + 1² n+1 Se dx = - ex cbc √x + a +1 -1 = In|x + al [cos(x) dx = sin(x) a [sin (ax) dx = cos(x) VECTOR PRODUCTS A-B = AB cose Ax B= ABsine
v=x0+at d = F --.cy enc 2= ² + 2q₂ (---) = ...ght EF F m m J = √F t = 4p p = my FSHF AE=W= K = P = a v=JF-dr P = F.v AU = mg h Xcm my :=FxP à-4-¹4 1-fr dm = 2mr¹ Σm.x, v = r@ K = =rxi = lo @= 0 + at 0=00 +00 + MECHANICS Universal gas constant, Boltzmann's constant, UNIT SYMBOLS J ADVANCED PLACEMENT PHYSICS C EQUATIONS a acceleration E- ene y F=_orce 1 J С m, ulse - kinetic e ey k spring constant { = len ul Langu. ir...mentu... m n m = mass P =, 0:cr P P otatio ali etia a 4 1 - Li...e J=, otential energy = elocity or s, cea W work done on a > stem xosition = coeficient of friction e-angle torque a gulur speed a gulur acceleration pase a gle mo...ent m = d.sta ce T is eioa F = -kax U. ---K(Ax)²2 J => max COS(col + o) -2-7 T= T₁ = 2₁ I2 |FG|-Gmm₂ Proton mass, m Net ron... ss, m = 1.67 x 10 kg Electron mass, m, = 9.11 10-31 kg A og du's numbe, N = 5.02 x 1023 mol-1 PREFIXES Factor Prefix. Symbol 109 giga G 106 mega M k 10³ kilo ₁-2 3 16 4 1. centi milli micro nano 10-12 pico = 2r, Ug = R= 8.31 J/(mol-K) kg = 1.38 x 10-23 J/K 1 unified atomic mass unit, Planck's constant, Gm,₂ r 1.67 x 10 27 kg meter, m mole, mol kilogram, kg hertz, Hz second, newton, N ampere, A pascal, Pa kelvin, K joule, J S E= . E-dA=2 %0 0° 0 1 0 E = _dV x AV = -√ E-dr V = 424/4 ΔΙ AV= U₂ =q' = - I 1 1992 4neo 2 c=xed C - EC - ELECTRICITY AND MAGNETISM A = area B magnetic field C = capacitance distance E-electric field = dQ dt لم = F 1 = Uc - eAv=C(AV)² |R-P = 1 = Nev A ΔΙ 1 9192 4² 7 R-ER Ε-ΣΕ P = 1AV ADVANCED PLACEMENT PHYSICS C TABLE OF INFORMATION NINIS AND CONVERN FORS Electron charge na gn..ude, watt, coulomb, volt, ohm, henry, - E emf .- force W С с V Ω H .7 1/2 3/5 1 = current J = current density Linductance . = length n = number of loops of wire per unit length N P = power wer number of charge carriers per unit volume Q = charge Ho = 4 x 10 (T-m)/A (T-m)/A 4J √2/2 √3/2 4/5 √2/2 √3/3 3/4 = point c¹ arge resistance 1 u = 1.66 x 10-27 kg = 931 MeV/c² h = 6.63 x 10-34 Js = 4.14 x 10-¹5 eV-s h: = 1.99 x 10-25 J-m = 1.24 x 10³ eV nm 8 = 8.85 x 10-12 C²/(N-m²) Coulomb's law constant, k = 1/(4reo) = 9.0 x 10° (N m²)/c² Vac um permittiv t,, Vacuum permeability. Magnetic constant, k = Ho/(4m) = 1x 10-7 1 atmosphere pressure, R= R- r = radius or distance t= time J = potential or stored energy ✓= electric potential v = velocity or speed p = resistivity - flux x= dielectric constant Fy=qvxB b.dl= Hol 150 x 119- e cron ol, 1 V-1.60 x 10 J Sd of light, Universal gravitational constant, Acceleratio a due to gravity at Earth s su..ace, The following assumptions are used in this exam. I. Ho Idex? dB-e 1 dbx F-SI dèxB B₂ = Honl = [Bau 8=E-di- 8=-L dt U₂ - 14² 1 atm = 1.0 × 10 N/m² = 1.0 x10³ Pa c = ..00 x 10 m/s G = .7 x 10" (N-m²j/kg² 8 = 9.8 m/s² farad, tesla. degree Celsius, electron volt, VALUES OF TRIGONOMETRIC FUNCTIONS FOR COMMON ANGLES 30° 90° 1 0 aine cose tane 0 - 60 J. 4/5 √3/2 3/5 1/2 1 4/3 √√3 do dt F T C eV 00 The frame of reference of any problem is inertial unless otherwise stated. II. The direction of current is the direction in which positive charges would drift. III. The electric potential is zero at an infinite distance from an isolated point charge. IV. All batteries and meters are ideal unless otherwise stated. V. Edge effects for the electric field of a parallel plate capacitor are negligible unless otherwise stated.
Transcribed Image Text:v=x0+at d = F --.cy enc 2= ² + 2q₂ (---) = ...ght EF F m m J = √F t = 4p p = my FSHF AE=W= K = P = a v=JF-dr P = F.v AU = mg h Xcm my :=FxP à-4-¹4 1-fr dm = 2mr¹ Σm.x, v = r@ K = =rxi = lo @= 0 + at 0=00 +00 + MECHANICS Universal gas constant, Boltzmann's constant, UNIT SYMBOLS J ADVANCED PLACEMENT PHYSICS C EQUATIONS a acceleration E- ene y F=_orce 1 J С m, ulse - kinetic e ey k spring constant { = len ul Langu. ir...mentu... m n m = mass P =, 0:cr P P otatio ali etia a 4 1 - Li...e J=, otential energy = elocity or s, cea W work done on a > stem xosition = coeficient of friction e-angle torque a gulur speed a gulur acceleration pase a gle mo...ent m = d.sta ce T is eioa F = -kax U. ---K(Ax)²2 J => max COS(col + o) -2-7 T= T₁ = 2₁ I2 |FG|-Gmm₂ Proton mass, m Net ron... ss, m = 1.67 x 10 kg Electron mass, m, = 9.11 10-31 kg A og du's numbe, N = 5.02 x 1023 mol-1 PREFIXES Factor Prefix. Symbol 109 giga G 106 mega M k 10³ kilo ₁-2 3 16 4 1. centi milli micro nano 10-12 pico = 2r, Ug = R= 8.31 J/(mol-K) kg = 1.38 x 10-23 J/K 1 unified atomic mass unit, Planck's constant, Gm,₂ r 1.67 x 10 27 kg meter, m mole, mol kilogram, kg hertz, Hz second, newton, N ampere, A pascal, Pa kelvin, K joule, J S E= . E-dA=2 %0 0° 0 1 0 E = _dV x AV = -√ E-dr V = 424/4 ΔΙ AV= U₂ =q' = - I 1 1992 4neo 2 c=xed C - EC - ELECTRICITY AND MAGNETISM A = area B magnetic field C = capacitance distance E-electric field = dQ dt لم = F 1 = Uc - eAv=C(AV)² |R-P = 1 = Nev A ΔΙ 1 9192 4² 7 R-ER Ε-ΣΕ P = 1AV ADVANCED PLACEMENT PHYSICS C TABLE OF INFORMATION NINIS AND CONVERN FORS Electron charge na gn..ude, watt, coulomb, volt, ohm, henry, - E emf .- force W С с V Ω H .7 1/2 3/5 1 = current J = current density Linductance . = length n = number of loops of wire per unit length N P = power wer number of charge carriers per unit volume Q = charge Ho = 4 x 10 (T-m)/A (T-m)/A 4J √2/2 √3/2 4/5 √2/2 √3/3 3/4 = point c¹ arge resistance 1 u = 1.66 x 10-27 kg = 931 MeV/c² h = 6.63 x 10-34 Js = 4.14 x 10-¹5 eV-s h: = 1.99 x 10-25 J-m = 1.24 x 10³ eV nm 8 = 8.85 x 10-12 C²/(N-m²) Coulomb's law constant, k = 1/(4reo) = 9.0 x 10° (N m²)/c² Vac um permittiv t,, Vacuum permeability. Magnetic constant, k = Ho/(4m) = 1x 10-7 1 atmosphere pressure, R= R- r = radius or distance t= time J = potential or stored energy ✓= electric potential v = velocity or speed p = resistivity - flux x= dielectric constant Fy=qvxB b.dl= Hol 150 x 119- e cron ol, 1 V-1.60 x 10 J Sd of light, Universal gravitational constant, Acceleratio a due to gravity at Earth s su..ace, The following assumptions are used in this exam. I. Ho Idex? dB-e 1 dbx F-SI dèxB B₂ = Honl = [Bau 8=E-di- 8=-L dt U₂ - 14² 1 atm = 1.0 × 10 N/m² = 1.0 x10³ Pa c = ..00 x 10 m/s G = .7 x 10" (N-m²j/kg² 8 = 9.8 m/s² farad, tesla. degree Celsius, electron volt, VALUES OF TRIGONOMETRIC FUNCTIONS FOR COMMON ANGLES 30° 90° 1 0 aine cose tane 0 - 60 J. 4/5 √3/2 3/5 1/2 1 4/3 √√3 do dt F T C eV 00 The frame of reference of any problem is inertial unless otherwise stated. II. The direction of current is the direction in which positive charges would drift. III. The electric potential is zero at an infinite distance from an isolated point charge. IV. All batteries and meters are ideal unless otherwise stated. V. Edge effects for the electric field of a parallel plate capacitor are negligible unless otherwise stated.
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