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Transcript

Physics for CCEA A2 Level Colourpoint Educational Rewarding Learning Pat Carson and Roy White Contents Unit 4 (A2 1): Momentum, Thermal Physics, Circular Motion, Oscillations, Atomic and Nuclear Physics 4.1 Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 4.2 Thermal Physics. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 4.3 Uniform Circular Motion. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 4.4 Simple Harmonic Motion. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 4.5 The Nucleus. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 4.6 Nuclear Decay. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 4.7 Nuclear Energy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 4.8 Nuclear Fission. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 4.9 Nuclear Fusion. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 Unit 5 (A2 2): Fields and their Applications 5.1 Force Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 5.2 Gravitational Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104 5.3 Electric Fields. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 5.4 Capacitors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 5.5 Magnetic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141 5.6 Deflection of Charged Particles in Electric and Magnetic Fields. . . . . . . . . . 163 5.7 Particle Accelerators. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182 5.8 Fundamental Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200 Unit 6 (A2 3): Practical Techniques 6.1 Measuring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211 6.2 Precision, Accuracy and Errors. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221 6.3 Analysis and Interpretation of Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224 Unit 7: Synoptic Assessment 7.1 Synoptic Assessment. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235 3 42a4.4 a = –Harmonic ω x Simple Motion 2 a = – ω2x 42a Students should be 42b able to: € ω = 2πf 4.4.1 Define simple harmonic motion using the equation a = 42b – ω 2 x where ω = 2πf ; 42a € 4.4.2 Perform calculations using x = A cos ωt ; 42c € 4.4.3 Demonstrate an understanding of s.h.m. measuring velocity x = Afrom cos the ωt ω =42c 2πf 42b graphs to include € € gradient of a displacement time graph; 42d x € 4.4.4 Know and be able to use the terms free vibrations, forced vibrations, resonance and x x = 42d A cos ωt 42c damping in this context; € € 2 4.4.5 Understand 42e the concepts ofωlight damping, overdamping and critical damping; € 42d x 42e 4.4.6 Describe mechanical examples of resonance € and damping; € 42f € ω2 F = ma = – mω 2 x 2 ω 42f point will oscillate 42ea € F = maabout = –itsmω 2 x Any object that is initially displaced slightly from stable equilibrium € equilibrium position. It will, in general, experience a restoring force that is proportional to its –ω 2 42g € displacement from the equilibrium point. 42f F = 42g ma = – mω 2 x–ω 2 € It consists of a € A simple pendulum is a system that behaves in this way. mass, called the bob, 2 attached to a length of42h string. –mω € Pendulum at rest Pendulum oscillating When hanging vertically the 2 2 –ω 42g –mω 42h resultant force on the bob is € € zero, the tension in the string is 1 2π f = = € equal and opposite to 42i the weight T 42hω Tension 1 2π –mω 2 of the bob. When the pendulum in string € 42i € f = = T ω bob is moved a little to one side and released it oscillates θ = ωt 42j 1 2π € Weight of 42i € backwards and forwards. The f = = the bob θ = ωt 42j T ω resultant force always acts This component of the bob’s € weight provides a restoring towards the equilibrium position x force towards the equilibrium point. € Then : cos θ = cos ωt = so the bob will vibrate around θ =OPωt 42j this point. € Then : cos θ x = cos ωt = € O A 42kωt But, since x OP = OA = am gives : x = A cos Defining Simple x = Rearranging – ω 2(SHM) 42aHarmonic aMotion € Then : cos θ = cos ωt = OP gives : x = A c ω = 2πfmotion if Rearranging A particle moves42b with€simple harmonic 42kωt from a fixed But,point sinceand OP = OA = amplitude, A, then : co 1 its acceleration42b is proportional to=its displacement x2πf = A cos 42L ω € 2 the direction 42c of the€ acceleration point. x =isAalways cos ωttowards that fixedRearranging gives : xx == A A cos cos ωt ωt 42L € € The defining equation for simple harmonic motion is: 42m ω x 0 sin ωt x = vA = cos–ωt 42c 2 € € x a = – ω x 42d v = – ω x 0 sin ωt x = 42m A cos ωt 42L € € € in ms–2 where a is the acceleration 2 A cos ωt – ω2x = – ω x is athe= displacement 42d 42n in m ω = 2πf € € 2 –2 € in s ω is a constant 42e € 42m v = 42n – ω x 0 sin ωt a = – ω 2 x = – ω 2 A cos ω € € 2 ω 42e x = A cos ωt€ € 2 € 49 F = ma = – mω 42f a = – ω 2 x = – ω 2 A cos ωt 42n x € € € F = ma = – mω 2 x 42f x € 42a 42k 42a 42b 42c 42d a = –But, ω 2 xsince OP = OA = amplitude, A, then : cos θ = 42c € x = A cos ωt Physics for ccea a2 level • UNIT 4 42d x € The minus indicates that the acceleration and displacement are in opposite directions. Newton’s 2nd law tells that a resultant force F is needed to produce this acceleration. 2 ω force and displacement. 42e € acceleration, The graphs below illustrate these relationships between a = – ω2x 42a a = – ω2x 42a 42f € a = – ω2x 42a 42b Acceleration, a ω = 2πf 42g €Gradient is –ω2 +ω2 Α € ω = 2πf 42b € x = A cos ωt 42c 42h € € x = A cos ωt 42c € +Α Displacement, x x 42d € 42i € –Α x 42d –Α € 42b 42e € 42f 42g € F = ma = – mω 2 x € –ωis2–mω2 Gradient 42c € –mω 2 42d € ω = 2πf Force, F +mω2 Α x = A cos ωt x +Α 1 2π f = = 42e € T ωω 2 Displacement, x ω2 ω2 42e € 42j FΑ = ma = – mω x –ω F = ma = – mω 2 x 42f € 2 € € –mω2 Α –ω 2 42g € : cos θ = cos ωt = Then 2 –ω 2 € 42gGradient is –ω 2 € F = ma = – mω 2 x θ = ωt 42f € x OP 42k 2 But, since isOP–mω = OA = amplitude, A, then : c 42h Gradient € Rearranging gives : x = A cos ωt –mω 2 42h 2 The amplitude displacement of the motion is denoted by the letter A. € 42hor maximum–mω 1 2π € f = velocity = € In simple harmonic motion the object’s displacement from the42i equilibrium position, T ω and x =that A cos ωt each of these 42L acceleration all vary with1 time. 2π To derive the appropriate equations describe 42i €quantitiesfit =is necessary = 1to return 2π to circular€motion. physical 42i € T f =ω = T ω θ = ωt 42j 42m v =€ – ω x 0 sin ωt Consider a point P moving in circle of radius A with € P θ = ωt θ = ωt 42j velocity angular x € 42j ω€. € Then : cos θ = cos ωt = The projection of P onto the diameter XY is the 42n point a = – ω 2 x = –Aω 2 A cos ωt OP R. As P moves in a circle starting at Y then to X and € x 42k But, since OP = OA = amplitude, € to Y,€the Then : cos θ =: cos cos =cos ωt = x finally back point R moves along Then θωt=the OP OP diameter from Y to X and back to Y. € X x x x R gives :Y x = A cos ωt O Rearranging 42k But, since OP = OA = amplitude, A, then : cos θ = The point R42k moves along the diameter with=simple But, since OP OA = amplitude, A, then : cos θ = A A harmonic motion. Rearranging gives : xgives = A: cos Rearranging x =ωtA cos ωt42L x = A cos ωt € 42L 50 42m 42n € x =ωtA cos ωt 42L x = A cos € 42m € 42m v = – ωvx 0=sin– ω ωtx 0 sin ωt € 42n 2 2 2 x = – ωωt A cos ωt ωω 42n a = – ωa2 x= = – – A cos € € v = – ω x 0 sin ωt a = – ω 2 x = – ω 2 A cos ωt 42c 42d €42c 42g€ € x = A cos ωt x –ω = 2A cos ωt 4.4 simple harmonic motion x €42d The periodic T is the time it takes R to complete one oscillation. This is of course the same as 2 x –mωtime 42h€ € it takes P to complete one circle. As with all vibrations, the frequency, f, is the number of vibrations made per second. ω2 42e € 42e € 42i € ω2 1 2π f = = F = ma = –Tmω 2 xω 42f € F = ma = – mω 2 x where T is measured in s 42f € ω is measured in rad s–1 θ = ωt 42j € f is measured in Hz –ω 2 42g € 2 –ω 42g € x R from the displacement of the point € The maximum 2 Then : cos θ = cos ωt = –mω 42h centre of oscillation O is called the amplitude. In this OP € 2 –mω 42h case it equals the radius of the circle OY or OP and is x 42k€ denoted But, since OP = OA = amplitude, A, then : cos θ = by A. The displacement x of R from O varies A 1 2π with time. Rearranging gives : x = A cos ωt 42i € f = = 1 Tf =at time ω =t =2π0 R is at Y, i.e. at extremity of the 42i € Suppose X T ω 42j oscillation. 42L €42j x 42m 42n €42L €42m € Y x = A cos ωt θ = ωt further at time t, P has moved through an € Suppose angle θ = ωt . At this time the displacement of R from € O is x. 42m v = – ωA sin ωt € x € Then : cos θ = cos ωt = OP= x € Then : cos θ = cos ωt OP x a = – ω 2 x = – ω 2 A cos ωt 42n 42k € But, since OP = OA = amplitude, A, then : cos θ = A x 42k But, since OP = OA = amplitude, A, then : cos θ = A Rearranging gives : x = A cos ωt Rearranging gives : x = A cos ωt € This is the displacement equation for SHM. 42L R θ O P x = A cos ωt x = A cos ωt v = – ωA sin ωtvelocity and acceleration during Simple Harmonic Motion Displacement, v = – ω x 0 sin ωt € In the diagram below the object R moves along the line YX with simple harmonic motion. The centre of oscillation is the point O. At time t = 0 the object is at Y and moving to towards O. a = – ω 2 x = – ω 2 A cos ωt The graphs theωt displacement x, (represented by the vector OR), the velocity v €42n – ω 2how A cos a = overleaf – ω 2 x =show € and the acceleration a of the object vary with time t. The time to complete one oscillation is the period T. € € Direction of motion of R X O Y R 51 t=T Displacement : cos= θ –=mω cosx ωt = FThen = ma 42f € € OP = 2πf F = ma = – mω 2 x 42f x x = since A cosOP ωt 42c € = OA Physics 42k € But, = amplitude, A, a2 thenlevel : cos•θ UNIT = 4 for ccea 2 A –ω 42g € Rearranging gives : x = A cos ωt = A cos ωt The displacement T T 3T The –ω 2 42gx varies sinusoidally. t=0 x 42d t=_ t=_ t=_ € 4 2 4 € value at any instant is given by –mω 2 42h € x = A cos ωt 42L € –mω 2 42h ω2 42e € € where A is the amplitude of the oscillation 1 2π 42i fv == – ω=x sin ωt 42m € 0ω T € 2 1 2π F = ma = 42f = €– mω x f = the displacement-time graph, we € Note that given42i ω can obtain the velocity at any instantT by drawing 2 2 x = –atωthe Apoint cos ωt – ωtangent 42n the2appropriate and finding x θa =2 ωt = ma42j = –€mω –ω 42g the gradient. € θ = ωt 42j € 2 x € Then2: cos θ = cos ωt = –mω 42h OP € x € Then : cos θ = cos ωt = x But, since OP = OA = amplitude, A, then : cosOP θ = 2 42k ω A x 1 2π 42k But, since OP = OA = amplitude, : cos θ 3T = 42i € f = = gives : x = A cos ωt Rearranging T A, then_ T _ _ t = 0 t= t= t= A T any instant ω The velocity at is equal to the 4 2 4 Rearranging gives : x = A cos ωt graph at that 1 2π gradient of the displacement-time = instant. While the general equation for v is not x = A cos ωt T 42L ω ωt CCEA specification, it turns out 42j € requiredθby= the € to be: x = A cos ωt 42L € = ωt 42m v = – ωA sin ωt x € € Then : cos θ = cos ωt = The velocity has a maximum value at the 42m ωt v = of – OPωA sin € instant the object through the centre of 2 x x– ω 2passes xOP = =– ω ωt 42n 42kθ €= oscillation. OAA =cosamplitude, A, then : cos θ = en : cos cos ωt aBut, == since A OP xx = – ω 2 A cos ωt – ω 2ωt 42n gives : x a= =A cos Rearranging , since OP = OA = amplitude,€A, then : cos θ = A € arranging gives : x = A cos ωt x = A cos ωt 42L € € Time Velocity t=T instant. = – ω x 0 sin ωt 42n € a = – ω 2 x = – ω 2 A cos ωt The acceleration has a maximum value of at the ωtinstant the object reaches the = – ω x = – ω 2 A cos € extremities of its oscillation. The acceleration is zero when the object reaches the centre of the oscillation. t=0 T t=_ 4 T t=_ 2 _ t = 3T 4 t=T Acceleration = A cos ωt The acceleration at any instant is equal to the 42m v = – ω x 0 sin ωt € gradient of the velocity-time graph at that Time 2 The minus sign tells us that the acceleration is always in the opposite direction to the displacement from O. 52 Time x But, since OP = OA = amplitude, A, then : cos θ = 1 2π A θf == ωt 42j 42i4.4€ = motion €simple harmonic Rearranging gives : x = A cos ωt T ω 42k The specification makes it clear that candidates are expected to know how xto perform calculations Then : cos θ = cos ωt = θ = ωt 42j € for displacement is based on time tOP = 0 occurring at the based x = A cos ωt . This expression € € point of maximum positive displacement. For the sake of completeness, the equations below show x 42k OA = these amplitude, A, then : cos θ = how the displacement and acceleration changeBut, withsince time OP and =also how equations would A x moving in the 42m ω x 0t sin changevif = the–time = 0 ωt occurred at € the equilibrium position with the particle Then : cos θ = cos ωt = Rearranging gives : x = A cos ωt € direction of positive displacement. There is no requirement to know the equation showing how OP velocity changes with time. x 42L 42n € 43a € 43b € 42k cos ωt a = – ω 2 x = – ω 2 A 42L € x = A sin ωt But, since OP = OA = amplitude, A, then : cos θ = A x = A cos ωt Rearranging gives : x = A cos ωt Based on time t = 0 at x = A 42m 43a v = – ωx x=0 sin ωt ωt A sin € x = A cos ωt 42L Displacement, ωt = – ωx22€ x x = A sin ωt a = – ω 22 A sin 43a – 2ωA2 sin A cos ωt ωt = – ω2x 43b a = – ωa2 x= =– ω € € 42m v = – ω x sin ωt € a = – ω 2 A sin ωt0 = – ω 2 x Acceleration, 42na 43c € 43a 43b 43d 43c 43e 43f 43d 43e 43g 43f x = A cos ωt 43b € 22 2 x 22 == AAsin cos ωt2 ωt x = A cos ωt € 43c € 2 cos ωt a = – ω 22x = –2 ω 2 A 42n Velocity and Displacement € x = A cos ωt x = A cos ωt € 2 sin 43c 2 va = =the––velocity ωA ωt A sin ωt = – ωdisplacement x x 2 = A2 from ω How does varies with the centre of oscillation? cos2 ωt € 2 2 2 2 2 2 2 2 v = ω A sin ωt = – ωA ωt time. The equations below show how the€displacement and vvelocity varysin with 43d € 2 2 2 2 x = A cos ωt v = – ωA sin vωt = ω A sin ωt € 2 2 2 2 43d € 2 2 θ= +A2cos θ ωt = 1 sin x cos v 2 = ω 2 A2 sin 2 ωt € The relationship between sine and cosine 43etells us that sin 2θ + cos 2θ = 1 . So, € 22 v = – ωA sin ωt sin43e ωt + cos 22 ωtsin =2θ1 + cos 2θ = 1 € € 2 2 2 v 222 A =22 ω ω sinA22 ωtsin+ ωt ω 22 A22 cos22 ωt = ω 22 A22 sin 2 ωt + cos 2 ωt = 1 43f € 2 2 2 2 2 2 2 A 2 sin 2 ωt Substituting for x and v this into the above gives: sinω ωt + cos 2 ωt+ =ω 1A cos ωt = ω A € 2 43f 22 2 +22 cos θ ω = 22 1A22 vsin +θ ω x 22 = ω 2 A2 sin 2 ωt + ω 2 A2 cos2 ωt = ω 2 A2 € € So : v 22 = ω 22 A22 – ω 22 x 22 and hence : v 2 + ω 2 x 2 = ω 2 A2 € 2 2 2 2 2 22 2 ωt 22 + cos 2 ωt 2= 1 sin – x v = ± ω A 43g ( ) v + ω 2 x 2 = So ω 2 :Av2 = ω A – ω x and hence : € € 2 2 2 2 2 ω 2 A2 sin 2 ωt + ω 2 A2 cos2 ωtSo=: ω v 2 A= ω 2 Av2 =– ω and 43g –hence x2 : ± ωx A ( 43h 43g 43i 43j 43h 43k 43i 43L 43j ) – x ) i.e. to left or to the right or up or = ± ω or( A The ± indicates the ωt velocity can bevpositive negative v = – ωthat A sin 2 2 2 2 € down, inv 2other words it refers to direction. The equation above is not required by the +ω x = ω A € specification, but it is so useful that it is43h v = ω Asake sin of ωtcompleteness. reproduced here for–the €: So : v 2 =22 ω 2 A2 – ω 2 x 2 and hence ωt a = – ω A cos43h v = – ω A sin ωt € v = ± ω ( A2 – x 2 )€ a = – ω 2 A cos ωt 43i € ±ωA a = – ω 2 A cos ωt 43i € € v = – ω A sin ωt ±ωA 43j € € 2 ω 2A ±ωA 43j € € 2 a = – ω A cos ωt ω 2A 43k € € ω 2A v = ± ω ( A22 43k – x 22 ) € € ±ωA € v = ± ω ( A2 – x 2 ) 43L € 2 2 53 v 43g = – ω A sin ωtSo : v = ω A – ω x and hence : 43h x Then : cos θ = cos ωt € = 2 2 2 ±2ω ( A – x ) = – ω A sin ωt 43h v 2 +OP ω 2 x 2 =v ω=43i A x 2v)ωt v = ±=ω– (ωA2 2A–cos a v = a2 – ωlevel A sin •ωtUNIT € 43hPhysics € for ccea € 4 2 x : € = OA 2 2 x 2 and hence v 2 = ωa 2 =A –ω ω 43gOP A: cos 43i So 42k But, since = :amplitude, A, –then cos ωt θ = € A 2 2 v = – ω A sin ωt 43h a = – ω 2 A cos ωt 43i – x v = ± ω A 2 Summary of EquationsRearranging for Displacement, Velocity and Acceleration ( ) ±ωA 43j gives : xa == A cos ωt v = – ω A sin ωt 43h € A cos ωt – ω 43i € € € € ±ωA 43j Displacement Acceleration € Velocity 2 Variation a ωt =43k – ω A cosa ωt 43i ω 2–Aω€2 A cos ωt±ωA x = A43j cos ωt 42L with v =€ – ω A sin 43h =43j 43i ±ωA € time € € €€ ω 2A 43k €At fixed point, At extreme displacement, Maximum value Amplitude = A 2 2 = ±ωA 43j 43k Max velocity maximum acceleration =2 ω A2 2 42m v = 43k –43i ω x 0 sin ωt ωa A=€ – ω A cos ±ωA 43j ωt € v = ± ω A – x ) 43L ( € € € €€ At fixed point, At extreme displacement, 2 2 Minimum value v = ± ω ( AAt fixed – x )point, acceleration = o 43Lminimum € displacement = o velocity = 0 2 ω A2 v = ± ω ( A2 – x 2 ) ±ωA 43j A cos ωtω A243k a = –43L ω 2 x = – ω 2 43k 42n with ω 2 A43L € 2 Variation € v = ± – x € € x a = – ω 43m ( ) € € € displacement a = – ω2x 43m €2 2 v = ± ω ( A2 v– x=43m 43L ω A€ 2 43k ) ± ω 2π(A2 – x 2a) = – ω 2 x 43L € € € a = –ω x 43m Other Equations 43n € ω =€ = 2πf € T 2π Angular velocity (sometimes called angular € 43nfrequency): ω = = 2πf 1 1 2 2 T =2 = = 0.2 seconds T 2π v = ± ω ( A43m –=x –) ω 244a 43L € x a 43m f 5 a =43n – ω€x 2π ω = = 2πf € 43n € ω = = 2πf € ω = 2πf = 2π × 5 = 10π T = 31.4 rad s–1 1 1 T Worked Examples T = €= = 0.2 seconds 44a f 5 2 2π 2 2 2π a€=€ – ω x ω ω= = 2πf 44b 43m a == ω31.4 x =rad(10π Example 1 == 2π × 5 = 10π s–1 ) × 5 ×10 –2 = 5π 2 = 49 43n 2πf € € 43n ω = = 2πf € T A steel strip is clamped at one end and made to vibrate at a frequency of 5 Hz T and € with an € 2 2 2 amplitude of 50 mm. x =–2A=cos 50 cos 44b a = ω x 44c = (10π ) × 5 ×10 5πωt == 49.3 mm(10π s–2 × 0.04 ) = 15 € € 2π 1 1 43n ω = 2πfω. (a) (i) Calculate the period, and velocity, T T=€ = the =angular 0.2 = seconds 44a f 5 T € = A ωt mm 44c x € = A cos ωt = 50 cos (10π × x0.04 = 15.5 ) cos 1 1 2π × 5 = 10π =€ 31.4 rad s–1 € Solution T = = ω ==2πf0.2= seconds 33 = 50 cos 10πt 44a 1f 15 33 T = = = 20.2 seconds –1 cos 10πt = = 0.660 x 44d = A–2cos ωt 2 –2 rad s 2 44a ω44b= 2πf = 5a 2π × x5 ==(10π 10π = 31.4 f = ω × 5 ×10 = 5π = 49.3 mm s ) 50 € € € 33 = 50 cos 10πt 10πt = cos–1 0.660 = 0.850 ω the = 2πf = 2π × 5at=maximum 10π = 31.4 rad s–1 (ii) Calculate acceleration displacement. 33 cos 10πt = = 0.660 0.850 44d = 0.027 s t = 44c x = A cos 10π × 0.04 = 15.5 mm 50–2 2 ωt = 50 cos 2 –2 ( 2 ) 10π € 44b a = ω x = 10π × 5 ×10 = 5π = 49.3 mm s –1 ( ) Solution 10πt = cos 0.660 = 0.850 1€ 1 2 –2 2 T = = = 0.2 seconds 2 44b a = ω x = 10π × 5 ×10 = 5π = 49.3 mm s–2 = 0.027 s 0.850 ( ) 5 (b) f€ At time t = 0, the displacement x = isA+50 cos ωtmm. Calculate t = x = A cos ωt –1 10π € 44a 44b 44c € € € 44d € 44e 54 ω = 2πf = 2π × 5 = 10π€= 31.4 rad s 33 = 50 cos 10πt € at ωt time=t 50 = 0.04 –33 = 50 cos 10πt 44c (i) the displacement x = A cos cos s.10π × 0.04 = 15.5 mm 33 € 33 cos 10πt = = 0.660 44d cos 10πt = = – 0.660 44e = A cos ωt 2 x = A 2 ωt 50mm cos 50s–2 10π × 0.04 = 15.5 xmm a44c = ω 2 xSolution = (10π ) × 5 ×10 –2 cos = 5π ==49.3 50 € € –33 = 50 cos 10πt –1 10πt = cos–1 0.660 = 0.850 cos – 0.660 = 2.292 (ii) the times during the first period at which the distance from the fixed point10πt is 33= mm. x = A cos ωt 0.850 33 cos 10πt = = – 0.660 2.292 44e s = 0.027 t =mm = 0.073 s t = cos ωt = 50 cos (10π × 0.04 ) = 15.5 50 x = A€ Solution 10π 33 50cos cosωt 10πt x == A 10π –1 10πt = cos – 0.660 = 2.292 €quarter cycle, In the 1st In the 3rd quarter cycle, In the 4th quarter cycle, In the 2nd quarter cycle, 33 = 50 33cos 10πt the displacement is –33 the displacement is –33 2.292 =x =0.660 44d x = A cos ωtcos 10πt = A cos ωt = 0.073 s0.027 half t =half a period t =exactly + 0.1 mm exactly mm a 10π € 50 –33 = 50 cos 10πt 33 € in (0.1 s) after 44f the time period (0.1 s) after the 33 = 50 cos 10πt t = 0.127 s cos 10πt = 44d –1= 0.660 the first quarter cycle. time in the second 10πt = cos 33 50 0.660 33 = 0.850 quarter cycle. cos 10πt = = 0.660 cos 10πt = = – 0.660 44e t = 0.027 + 0.1 –1 50 50 = 0.850 44f € 10πt = cos 0.660 0.850 t = 0.073 + 0.1 –1 s € s = 2.292 t = 0.12744g t = 10πt ==cos0.027 10πt = cos–1 0.660 = 0.850 – 0.660 t = 0.173 s 10π 0.850 0.850 2.292 = 0.027 s t = t = = s0.073 s t == 0.027 10π t = 0.073 + 0.1 10π 10π 44g € 2π 2π t = 0.173 s € T = so : 1.5 = x calculations = A cos ωt above it is essential that the44h Note that in the calculator is setωto ω t = 0.027 + 0.1 x = A cos ωt € 2π 44f “radian mode” calculating the cosines and inverse cosines. Rearranging : ω = € x when == tA = 4.19 rad s–1 50 cosωt 10πt = cos 0.127 s –33 = 50 cos 10πt –33 2π 2π 1.5 T = so : 1.5 = € € ω ω 33 –33 = 50 33cos 10πt 44h cos 10πt = = – 0.660 cos 10πt = = – 0.660 44e 2π 50 t 50 = 0.073 + 0.1 = 4.19 rad s–1 Rearranging : ω = –1 33 44g € 44i 1.5 ±ωA = 4.19 × 0.018 = ± 0.075 ms = 2.292 cos 10πt = t = 0.173 = – 0.660 44e10πt = cos–1 – 0.660 € s –1 10πt = cos 50 – 0.660 = 2.292 2.292 = 0.073 s t = –1 –1 10π 2.292 10πt = cos – 0.660 = 2.292 ( ( ) ) 5.5 Magnetic Fields Students should be able to: 5.5.1 Explain the concept of a magnetic field; F = BI that there is a force on a current-carrying conductor in a perpendicular 5.5.2 Understand magnetic field and be able to predict the direction of the force; 5a 5a using the equation F = BI ; 5.5.3 Define magnetic flux density 5b € 5a F = BI φ = BA 5.5.4 Define the unit of magnetic flux density, the tesla; 5.5.5 Understand the concepts of magnetic flux and magnetic flux linkage; 5b 5c €€ φ = BA 5b for magnetic flux, φ = BA , and flux linkage, 5.5.6 Recall and use the equations € Nφ = BAN ; 5.5.7 Define the unit of magnetic flux, the weber; € 5d 5cexperimentally Nφ = BAN € Faraday’s 5.5.8 State, use and demonstrate Lenz’s laws of 5c Nφ =and BAN € electromagnetic induction; 5.5.9 Recall and calculate induced e.m.f. as rate of change of flux linkage with time; € € € I Ns V F = p for transformers; B = and use5athe equation s = 5.5.11 Recall Vp Np Is I 5.5.12 Explain power losses in transformers and of high voltage transmission of €€ the advantages FF 5e B = electricity; 5e B = I I € F = BI € 5d works; 5.5.10 Describe how a transformer 5d 5e = 4.5 ×10 –2 × 1.25 × 0.25 5f The space surrounding a magnet where a magnetic force is experienced is called a magnetic field. = BI BIof metals such as iron and steel 0.014is N €€ objects This magnetic=force experienced by other magnets, FF =made –2 and also by current carrying conductors. The direction of a magnetic field a1.25 point× is0.25 taken as 4.5×10 ×10 –2 at == 4.5 ×× 1.25 × 0.25 5f5f the direction of the force that acts on a north pole placed at that point.N = 0.014 5g € shape B The of a magnetic field can be represented by a magnetic field lines (lines of magnetic force). ⊥ Arrows on the lines show the direction of the magnetic force. Since a north pole is repelled from another north pole and attracted by a south pole the direction of a magnetic field is from North to 5g B €€ South. 5g B⊥ = 0.014 N 5h ⊥ € MagnadurBmagnets are commonly used in school laboratories to = produce a strong magnetic field. Two are arranged on a U-shaped steel yoke as shown in the diagram. The diagram€also shows the 5h B= € shape and direction of the magnetic5hfield that is created byBthis = € F = B⊥ I arrangement. N S 5i B⊥ in = the 0.2 central × cos 40 The field lines portion are parallel and uniformly € spaced this indicates a magnetic field in this region F = B € that is uniform, = 0.153 T F = B⊥⊥II i.e. it has the same strength at all points in the region. The field B⊥ = 0.2 × cos 40 F = 0.153 × 0.5 × 0.3 × lines cos 40 weakens at the edges as indicated by the increased spacingBbetween of force. ⊥ = 0.2the = 0.023 N 5j € –5 2.0 ×10 T 5i5i = 0.153 T = 0.153 T F = 0.153 × 0.5 × 0.3 F = 0.153 × 0.5 × 0.3 = 0.023 N = 0.023 N 141 Physics for ccea a2 level • UNIT 5 Electromagnetism A conductor carrying an electric current is surrounded by a magnetic field. Different shaped conductors produce magnetic fields of various shapes. The magnetic field lines due to the current in a straight wire are circles concentric with the wire. The shape can be revealed by sprinkling iron filings around the wire. The direction can be obtained by using the right hand screw rule. If a right handed screw moves forward in the same direction as the current, then the direction of rotation of the screw gives the direction of the magnetic field lines. Current carrying wire Iron filings Current Direction of magnetic field The direction can also be obtained using the right hand grip rule, this is illustrated in the diagram above. Pretend to grab the wire with your right hand, your thumb pointing in the direction of the current. Your fingers curl in the direction of the magnetic field. The diagram below shows a view looking along the current carrying wire. represents a current flowing into the page (away from you) and the symbol The symbol represents a current flowing out of the page (towards you). You should apply the right hand grip rule to the diagrams below to confirm the magnetic field direction in each case. Magnetix flux lines Current carrying wire with current direction into the page 142 Current carrying wire with current direction out of the page 5.5 Magnetic Fields Force on a Current in a Magnetic Field When a current flows through a conductor in a magnetic field the conductor experiences a force. This can be demonstrated using the apparatus shown. When a current is passed along the flexible wire, the wire moves up. The force acts at right angles to both the current and the magnetic field direction. Upward force on the current carrying wire Force on the current carrying conductor F N Magnetic field direction North to South S B I Conventional current direction, positive to negative The direction of the force (movement) of the wire is obtained from Fleming’s Left Hand rule. Extend the thumb, first and second fingers of the left hand so that they are at right angles to each as shown in the diagram above. The first finger represents the magnetic field direction. The magnetic field direction is from the north pole (N) to the south pole (S). The second finger represents the conventional current direction, from positive to negative. The thumb represents the force acting on the current carrying wire. This force arises because of the interaction of the two magnetic fields: the uniform field of the permanent magnet and the field due to the current in the wire. The magnetic field lines of force are vectors and the field lines due to two fields have to be combined vectorially. This interaction of magnetic fields is illustrated below and the resultant field is some times called a catapult field. The field lines resemble the stretched rubber band in a catapult, the conductor experiencing a force in the direction shown. S N Uniform field of the permanent magnet + = Magnetic field due to the current which is INTO the page S N The resultant magnetic field is sometimes known as a catapult field 143 5a 5b 5c 5d 5e 5f 5g 5h 5i 5a F = BI Physics for ccea a2 level • UNIT 5 F = BI When the current and € the magnetic field are in the same direction or act at 180o to each other 5b φ = BA the force is zero. It is a maximum when the current and the magnetic field are perpendicular to € each other. φ = BA In the case when the magnetic field and the current are at right angles the force is given by: € F 5c = BI € Nφ = BAN where B is the magnetic flux density in tesla (T) Nφ = BAN I is the current in the conductor in amperes (A) € φ 5d = BA € € is the length of the conductor in the magnetic field in metres (m) F = BI flux density is used to indicate the strength of a magnetic field. Magnetic flux The term magnetic measured in € units known as the € density is F tesla, symbol T. The tesla is defined from the equation Nφ BAN 5e = Re-arranging B = shown above. the equation gives: I F B φ= = BA I € The magnetic flux density the force per unit current carrying length. A current € is defined F =asBI € € carrying length is the product of the current and the length of the conductor in the magnetic field. –2 € = 4.5 × 1.25 × 0.25 5f =A BAN of and a length of 0.5 m is×10 a current carrying length of 1.0 A m. € A current F Nφ = 2.0 BI = 0.014 N € A magnetic field –2 density 1.0 T will exert a force of 1.0 N on a current carrying length of flux F of == 4.5 ×10 × 1.25 × 0.25 1.0 Am B when the current and magnetic field directions are perpendicular to each other. I N = 0.014 € € 5g B⊥ Worked Examples € F € Example B⊥=1 BI € –2 = 4.5wire, × 1.25 × 0.25 F×10of A straight B = €length 60Bcm, carries a current of 1.25 A. Calculate the value of the 5h = I on force that acts a length of 25 cm of the wire is placed at right angles to a = 0.014 N this wire when € magnetic field of flux density 4.5×10–2 T. B= Solution €€ B⊥F = BI € F = B⊥ I Uniform magnetic field of flux density 4.5 × 10-2 T acting into the page 12.5 A = 0.2 × cos 40 = 4.5 ×10 –2 × 1.25B×⊥ 0.25 F = B⊥ I = 0.153 T 5i = 0.014 N 25 cm B = 0.2 × cos 40 ⊥ € 0.153 × 0.5 × 0.3 Only B= 0.25 m of the wire is in Fthe=magnetic = 0.153 T field. = 0.023 N F = 0.153 × 0.5 × 0.3 € Application of Fleming’s Left Hand Rule tells B⊥ us that = the0.023 force N acts on the wire to the left. € F = B⊥ I € 2.0 ×10 –5 T 5j B⊥ 2= 0.2 × sin 50 Example € B = 0.153 –5 € A straight T wire a current of 500 mA is placed in a magnetic field of flux density T carrying 2.0 =×10 € €€ 144€ 0.2 T. length×of0.5 the×wire F The = 0.153 0.3 in the magnetic field is 30 cm and the wire makes an angle of € o 50 with the magnetic field direction. Calculate the force on the current carrying wire due = 0.023 N to the magnetic field. F = B⊥ I B⊥ = 0.2 × cos 40 = –50.153 T T 2.0 ×10 F = 0.153 × 0.5 × 0.3 = 0.023 N € 5g € B⊥ 5.5 Magnetic Fields € 5g 5h = 4.5 ×10 –2 × 1.25 × = 0.014 N 5f B⊥ € SolutionB 5g The magnetic flux density is resolved into a component perpendicular to€the wire B⊥ and a € 5hparallel to the wire, B= . Only the perpendicular component produces a force on component the wire.. = € F = B⊥ I € 50 B⊥ = 0.2 × sin F = = 0.153 T B⊥ = F = 0.153 × 0.5 × 0.3 = 5i = 0.023 N F = 5h 5i B€ = B⊥ I B= 50° 0.2 × cos 40 0.153 T 0.153 × 0.5 × 0.3 = 0.023 N 5j F = BI € B⊥ F = B⊥ I B⊥ = 0.2 × cos 40 = 0.153 T F = 0.153 × 0.5 × 0.3 5i € 2.0 ×10 Verification of –5F T = BI = 0.023 N € can be2.0 The apparatus used how the force acting on a current carrying T ×10to–5 investigate 5j shown below conductor depends on the current flowing in the conductor and the length of the conductor in the € magnetic field. € Conventional current direction 5j € 2.0 ×10 –5 T Magnet (south) € Magnet (north) Aluminium rod (clamps not shown) Electronic balance An aluminium rod (e.g. the upright from a retort stand) is clamped horizontally above a sensitive electronic balance. The rod is connected to a variable low voltage supply. An ammeter connected in series with both will allow the current to be measured. A permanent magnet is placed on the balance and the aluminium rod positioned so that it is located in the centre of the magnetic field. The balance is set to read zero after the magnets have been placed on it. When a current is then passed along the clamped aluminium rod the rod experiences a force due to the interaction of the permanent magnetic field and the magnetic field due to the current in the aluminium rod. In the case shown in the diagram, the force is upwards, use Fleming’s Left Hand Rule to verify this. Since the magnet exerts an upward force on the rod, then by Newton’s third law, the rod must exert an equal but opposite force on the magnets. This downward force will cause the reading on the electronic balance to increase. 145 The current is varied with a single magnet in place. This ensures that the length of the conductor in the magnetic field remains constant. A graph of the force (y–axis) against current (x–axis) produces a straight line through the origin. This is verification that the force on the current carrying conductor in the magnetic field is directly proportional to the current. Force / N Physics for ccea a2 level • UNIT 5 0 Current / A The current is then fixed and a number of identical magnets are placed side by side, this changes the length of the conductor in the magnetic field. A graph of force (y–axis) against the length of conductor in the magnetic field (x–axis) produces a straight line through the origin. This is verification that the force on the current carrying conductor in the magnetic field is directly proportional to the length of the conductor in the magnetic field. Force / N 0 0 0 Length of conductor in the magnetic field / m Exercise 33 1 The horizontal component of the Earth’s magnetic field has a flux density of 2.0×10–5 T. A straight piece of wire XY 1.2 m long of mass 0.8 g is resting on a wooden bench so that it is at right angles to the magnetic field direction. A current is passed through the wire which just causes the wire to lift off the bench. X Y (a) State the direction of the current in the wire. (b) Calculate the current. 2 A single-turn rectangular wire loop ABCD hangs from a sensitive balance so that its lower portion is in a region of uniform magnetic field. The direction of the magnetic field is at right angles to the plane of the loop. This arrangement is shown in the diagram. The upper portion of the loop is not in the magnetic field. When a current passes round the loop in the direction ABCD, the balance reading gets less. 146 Current leads and support from the balance Uniform magnetic field A D B C