Strand H Unit 3: Electromagnetic induction. Text Answers. Exercise H.3.1 Answers. a force of F = N. Since F = Bev,

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Dominick Morris
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1 Exercise H.3.1 Answers 1. The magnetic field B = 0.6T and the electron of charge C experiences a force of F = N. Since F = Bev, vv = FF BBBB = NN CC 0.6TT = 30000mmss ms/s = 30km/s= km/h 2. Q = +2µC = C B = 0.8T v = 10m/s FF = BBBBBB = 0.8TT CC 10mmss 1 = NN Using Fleming s Left hand Rule, the first finger into the page and the second finger in the direction of the charge velocity (since it s a positive charge) the thumb shows the force to be vertically upwards. 3. A. Cathode and anode providing the accelerating voltage B. Electron gun (the source of electrons) C. The external magnetic field D. The vacuum tube E. The screen (or fluorescent coating) 4. L = 20cm = 0.2m B = 2T t = 3s d = 20cm The change in magnetic flux ΦΦ= B A = B L d = 0.2T 0.2m 0.2m = Tm 2. From Faraday s Law, the emf (and therefore the voltage across the rod) is given by: VV = εε = ΔΦ Δtt = TTmm 2 = = 1.3mmmm 6ss From Fleming s Left hand rule, with the first finger into the page and the second finger in the opposite direction to the rod velocity (the direction of current) the thumb shows the force and therefore the acceleration of the electrons t to be

2 vertically downward. Thus end A becomes positive whilst end B becomes negative. L = 10cm = 0.1m B = 1.3T vrod = 0.6m/s t = 6s ΦΦ=? The magnetic flux ΦΦ= B A where A is the area swept out by the moving rod. since speed = distance / time dd = vvvv = 0.6mmss 1 6ss = 3.6mm The area swept out by the rod is therefore found by Then the magnetic flux is given by AA = LL dd = 0.1mm 3.6mm = 0.36mm 2 Φ = BBBB = 1.3TT 0.36mm 2 = 0.47TTmm 2 = 0.47WWWW where Wb is the Webber, the derived unit of magnetic flux. Since the magnetic flux was initially zero, this is also the change in magnetic flux. From Faraday s Law The circuit resistance is 2Ω. Thus εε = ΔΦ Δtt = 0.47WWWW = 78mmmm 6ss II = VV RR = VV 2ΩΩ = 0.04AA = 40mmmm

3 Exercise H Your answer should include at least 5 from the following; As the coil rotates under the influence of an external rotational force The rotating wire constitutes a movement of charge in the stationary magnetic field The rotating electrons in the wire coil generate an induced magnetic field This induced magnetic field combines vectorially with the external field The electrons experience a force perpendicular to their direction of motion and the external magnetic field The force causes acceleration of the electrons in the direction of the force, creating an induced current in the wire The direction of the electrons under rotation relative to the field switches every 180 of rotation The force experienced and hence direction of induced current therefore switches direction, creating an alternating current in the wire 2. At 90 as shown, with the coil rotating anticlockwise, length AB is coming out of the page. Therefore with the first finger in the direction of the field and the second finger out of the page the thumb shows the force on the electrons to be vertically upwards. The induced current is therefore vertically downward from A to B. 180 later the rotating electrons are going into the page. Since the direction of the current (the rotating electrons in the wire) have reversed relative to the field, so too does the force on the electrons. The induced current is therefore vertically upward, from B to A. 3. A w B z C x D y rpm = 20/60s = 1/3 of a revolution per second. The time period is thus 3s. Since frequency is 1/T, the frequency of rotation and hence induced voltage is 0.33Hz.

4 Challenge Question ε (V) Hz 50Hz t (s) Exercise H Vp = 120V Vs = 9V Ns = 60 Np =? From the transformer equation VV pp = NN pp. Rearranging to make Np the subject we VV ss NN ss obtain; NN pp = VV PPNN ss VV ss = 120VV 60 9VV = 800 tttttttttt 2. If the bulb operates normally, from P = IV the current through the bulb and therefore the current on the secondary coil is given by; II = PP VV SS = 60WW 230VV = 0.26AA Since we assume the transformer is 100% efficient, we can now use the transformer efficiency equation with the efficiency set to 1. therefore II SS VV SS II PP VV PP = 1 II PP = II SSVV SS VV PP = 0.26AA 230VV 12VV = 4.98AA

5 3. A transformer reduces energy loss by using coils of low resistance wire and by using a laminated iron core. This reduces secondary currents forming in transformer core that could dissipate energy through electron electron and electron ion core collisions (a heating effect called Joule heating) 4. Since the number of turns on the secondary coil is less that that on the primary, the voltage induced on the secondary will also be less. This is a step down transformer. Vp = 200V Vs =? Ns = 20 Np = 150 From the transformer equation VV SS = VV PP NN SS 200VV 20 = = 26.6VV NN PP 150 Challenge Question Vp = 240V Vs = 12V Ns =? Np = 1000 P = 40W Ip = 0.2A (a) The number of turns on the secondary coil is found using the transformer equation; NN ss = NN ppvv ss VV = = 50 tttttttttt VV pp 240VV (b) The current in the secondary coil is the same as the current through the heater. From P = IV; II = PP VV SS = 40WW 12VV = 3. 3 AA (c) From the transformer efficiency equation TTTTTTTTTTTTTTTTTTTTTT eeeeeeeeeeeeeeee = II SSVV SS VV = = 0.83 = 83% II PP VV PP 0.2AA 240VV

1 Fig. 3.1 shows the variation of the magnetic flux linkage with time t for a small generator. magnetic. flux linkage / Wb-turns 1.

1 Fig. 3.1 shows the variation of the magnetic flux linkage with time t for a small generator. 2 magnetic 1 flux linkage / 0 10 2 Wb-turns 1 2 5 10 15 t / 10 3 s Fig. 3.1 The generator has a flat coil