AQA A2 Physics P25 Electromagnetic Induction Kerboodle Answers

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25.1 Generating electricity AQA A2 Physics P25 Electromagnetic Induction Kerboodle Answers : Page No. 412

1 a When a bar magnet is pushed into the coil of insulated copper wire connected to a galvanometer, an induced current is set-up in the coil due to change of magnetic field through it. As a result, galvanometer gives a deflection (say towards left). 2 a Explain why the lamp lights when the weight descends. b What difference would have been made if the magnet had been much stronger c Explain why a lamp connected to a dynamo lights when the dynamo turns. d Why is the dynamo easier to turn when the lamp is disconnected 3 a i Downwards ii Use the right hand rule because the motion of the rod through the magnetic field (the dynamo principle) is generating an emf between its ends. The left hand rule is used to find the force acting on a wire (the motor principle) when there is already a current in it, and you place it in a field. In this case The magnetic field is S to N (first finger/pointer finger0 The motion is downwards (thumb) The direction of the induced emf is in the direction of the middle finger. The direction of the emf is always “conventional”, that is to say, the direction of the positive terminal. (And the direction a positive current would flow if there was a circuit.)

25.2 The laws of electromagnetic induction AQA A2 Physics P25 Electromagnetic Induction Kerboodle Answer : Page No. 416

1 a i 1.1 mWb ii 2.0 s iii 0.54 mV 2 a. 1.4 mWb b 23 mV 3 a: i 4.5 x 10^-4 m² ii the flux linkage through the coil. 1.5(4) mWb b i the change of flux linkage through the coil, 3.1 mWb ii the magnitude of the induced emf. 33 mV 4 a 8.0 μWb b 40μV

25.3 The alternating current generator AQA A2 Physics P25 Electromagnetic Induction Kerboodle Answer : Page No. 419

1 a Anac generator produces an alternating emf with a peak value of 8.0 V and a frequency of 20 Hz. Sketch a graph to show how the emf varies with time. b The frequency Of rotation Of the ac generator in a is increased to 30 Hz. Dn the same axes, sketch a graph to show how the emf varies with time at 30 Hz. 2 flux linkage 0, 0, -BAN; induced emf 0, -e0, 0 3 a 26mWb b i Show each side of the coil moves at a speed of 6.0 ms- ii Show that the peak voltage is 8.1 V. 4 a Load refers to the amount of current a machine/device/instrument draws. Ideally a motor at no load means, it does not provide any mechanical output which again means that the torque produced by the motor is 0(ideally). By law of conservation of energy the input power to the Motor = output power given by the shaft (ideally ie. neglecting losses) So when mechanical output power is zero the input power should also be zero. Which inturn means that the motor does not draw any current (if it draws current then the input power is non zero which means that the motor is on load) b In an electric motor, almost all types, the current is proportional to the torque. If the load increases, in other words, the torque increases and the current will increase.

25.4 Alternating current and power AQA A2 Physics P25 Electromagnetic Induction Kerboodle Answer : Page No. 422

1 a 5 ms b 0.0707 A 2 a i the rms current, 2.12 A ii the rms pd across the resistor, 8.49 v b i the peak power, 36W ii the mean power supplied to the resistor. 18 w 3: 4.35 A ii

6. 15 A

iii the peak power. 2.0 kW b The fuse melts only if therms current exceeds 5 A. The rms current in this case is less than 5 A, so the fuse does not melt. 4 a i the rms current through it, 4.0 A ii the peak Current through it, and 5.66 A iii the peak pd across it. 17.0 V b i Vrms =Irms R = 4.0 x 0.5 = 2.0 V ii 19.8 v    

25.5 Transformers AQA A2 Physics P25 Electromagnetic Induction Kerboodle Answer : Page No. 425

1 a 11.5V b On ideal transformers, you need to know that power in = power out.

• If you connect something to the output you increase power out.
• This increases power input.
• This increases input current.
• By applying the idea of conservation of energy.
• There is no energy lost in the core or windings.

2 a The magnetic flux produced by the primary coil should be equal to the magnetic flux passing through the secondary coil so as to increase the efficiency of the transformer. In principle the magnetic flux passing through the secondary cannot be greater than the magnetic flux produced by the primary coil. Transformers only vary the potential difference b Transformers operate on the principle that current in a wire generates a magnetic field. And more importantly that a change in magnetic flux creates a current in a wire within that field. The input side of a transformer consists of multiple wraps around an iron core. The outputs is a similar set of wraps around the same core (the ratio of number of input wraps to output directly determines how much the voltage is transformed.) Since electricity in the output is dependent on CHANGE in magnetic flux, a DC input will not create any output except at the time it is switched on or off. So AC current is used which will continuously provide changing magnetic flux and therefore power at the outputs.

a Calculate the primary voltage needed for a secondary voltage of 230″. b A ?30V 60 W lamp is connected to its secondary coil. Calculate the Current through i the secondary coil, ii the primary coil. State any assumptions made in this calculation. 4 a The primary reason that power is transmitted at high voltages is to increase efficiency. As electricity is transmitted over long distances, there are inherent energy losses along the way. High voltage transmission minimizes the amount of power lost as electricity flows from one location to the next. How? The higher the voltage, the lower the current. The lower the current, the lower the resistance losses in the conductors. And when resistance losses are low, energy losses are low also. Electrical engineers consider factors such as the power being transmitted and the distance required for transmission when determining the optimal transmission voltage. Banner 2

Practice questions: Page No. 426-429

1 (a) Ammeter deflects in one direction and then in the opposite direction. (b) (i) As it emered the coil, The acceleration of the magnet would be reduced…because the magnetic field produced by the current induced in the coil gives an upwards force on the N-pole of the magnet that opposes the movement of the N-pole towards the coil. (ii) As it left the coil. (4 marks) The acceleration of the magnet would again be reduced… because the magnetic field produced by the current induced in the coil gives an upwards force on the S-pole of the magnet that opposes the movement of the S-pole away from the coil. (c) Relevant points include:

• The magnet would now fall with an acceleration of g
• An emf would still be induced in the coil
• No current could be produced by the induced emf
• No induced magnetic field would be produced
• There would be no opposing forces on the magnet due to an induced current.

2 (i)

• Φ represents magnetic flux

(ii) The unit of Φ is weber (Wb), or tesla metre² (T m²). (b)  (i) is a maximum i.e. where the gradient of the B–t graph is greatest, ∴ draw tangent to line at t = 0.5 or 1.0 s (t = 0 is also acceptable). (iii) Pendulum would have to be shorter; ¼ of the original length. (iv) Explain why this increases the maximum induced ernf. The maximum speed of the magnet is increased ∴ the coil cuts the flux at a higher rate or rate of change of flux increases. (v) Possible alternative ways:

• use a stronger magnet
• use a coil with more turns
• use a coil of greater area
• place a soft iron core in the coil
• use a larger amplitude of oscillation of the magnet.

3 (a) Distance between pulses = 7.2 cm ∴time between pulses = 72 ms Time for 1 revolution of axle = 4 × 72 = 288 ms Number of revolutions per minute or 205−211 with an extended range of values of the time between pulses. (b) (i) Relevant points are:

• Movement of a magnet changes the magnetic flux through the coil (or changes the flux linked with the coil)
• The induced emf is proportional to the rate of change of flux linked with the coil (or an emf is produced because the magnetic flux cuts through the coil).

(ii) (c

• Direction of induced emf (or current) opposes the change that produces it.
• A pulse is produced as a magnet approaches and leaves a coil.
• Forces on a magnet repel it as it approaches and attract it as it leaves.
• Current therefore flows one way as the magnet approaches and the opposite way as it leaves.

(d) Waveform drawn on Figure 2 to show:

• larger peak voltages
• + and − peaks closer together
• narrower (sharper) peaks
• sets of peaks closer together.

4 (a) b (i) (ii) An emf is induced once the coil is rotating, because the coil cuts the magnetic flux continuously. This induced emf opposes the emf that is causing the coil to rotate, thereby reducing the current. (iii) 5 (a) Magnetic flux linkage is the product of magnetic flux and the number of turns that cut the flux or = NΦ (= NBA, where B is normal to A), with terms defined. The unit of NΦ is weber turn (Wb turn) (b

• The alternating voltage, Vin, supplied to the input causes an alternating current in the primary coil
• The ac in the primary coil produces an alternating magnetic flux in the core
• This alternating flux passes through the secondary coil
• The induced emf, Vout, in the secondary coil is proportional to the rate of change of flux linkage
• There are fewer turns on the secondary than on the primary, so Vout < Vin

(C) (i) . (ii) Reasons for imperfect efficiency:

• Thermal energy losses caused by the currents in the coils, which have resistance
• Energy is required to magnetise, demagnetise and re-magnetise the core
• Eddy currents induced in the core cause thermal energy losses in the core
• Imperfect flux linkage: not all the magnetic flux produced by the primary coil passes through the secondary coil.

6 (a) (i) (ii) (iii) If transformer is perfectly efficient, primary power = secondary power (or IpVp = IsVs) Power input = primary power = secondary power = power output = 8 × 30 = 240 W (b) Relevant points include:

• The pendulum bob cuts through the magnetic flux of the field of the magnet
• There is an induced emf in the bob
• This emf causes circulating currents (eddy currents) in the brass bob
• These currents cause thermal energy losses in the bob
• The energy converted in this way comes from the kinetic energy of the bob, causing it to slow down rapidly that is, heavy damping

7 (a) (i) (ii) (b)

• Power is lost from the cables (or pd is dropped along them)
• Longer cables have greater resistance
• Power lost from cables = Irms 2r
• Less power is wasted if a high voltage is used
• Because at high voltage the current needed is less (for the same power)
• ac voltages can be changed using transformers
• Transformers cannot be operated by steady dc.

8 (a)

• (i)

(ii) Banner 3

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