Monday, April 10, 2017

A student attempts to show the interference of light using two identical green LEDs.




Question 1
A student attempts to show the interference of light using two identical green LEDs.
Which statement explains why the experiment will not succeed?
A The light waves from the sources are not coherent.
B The light waves from the sources do not have the same amplitude.
C The light waves from the sources have a range of wavelengths.
D The light waves from the sources are not monochromatic.





Reference: Past Exam Paper – June 2014 Paper 13 Q28





Solution 1:
Answer: A.
Interference occurs when the light waves are coherent (that is, they have constant phase difference and the same frequency). So, if they light waves from the sources are not coherent, the experiment will not succeed.

Amplitude is not a condition for interference.

Since the LEDs are identical, they are monochromatic and have the same wavelength.

Friday, March 31, 2017

An ideal operational amplifier (op-amp) has infinite open-loop gain and infinite input resistance (impedance).







Question 1
(a) An ideal operational amplifier (op-amp) has infinite open-loop gain and infinite input resistance (impedance).
State three further properties of an ideal op-amp. [3]


(b) The circuit of Fig. 10.1 is used to detect changes in temperature.


Fig. 10.1

The voltmeter has infinite resistance.
The variation with temperature θ of the resistance R of the thermistor is shown in Fig. 10.2.


Fig. 10.2

(i) When the thermistor is at a temperature of 1.0 °C, the voltmeter reads +1.0 V.
Show that, for the thermistor at 1.0 °C, the potential at A is −0.20 V. [4]

(ii) The potential at A remains at −0.20 V.
Determine the voltmeter reading for a thermistor temperature of 15 °C. [2]

(c) The voltmeter reading for a thermistor temperature of 29 °C is 0.35 V.
(i) Assuming a linear change of voltmeter reading with change of temperature over the

(ii) Suggest why your answers in (b)(ii) and (c)(i) are not the same. [1]





Reference: Past Exam Paper – November 2014 Paper 41 & 42 Q10





Solution 1:
(a)
zero output resistance / impedance
infinite bandwidth
infinite slew rate


(b)
(i)
{from graph:} at 1.0 °C, thermistor resistance is 3.7 kΩ
{for inverting amplifier: voltage gain = Vout / Vin = - Rf / Rin}
amplifier gain = –R / 740 = –3700 / 740 (negative sign essential)
gain = –5.0      C1
{potential at A = Vin = Vout / gain = = 1.0 / –5.0}
potential at A = 1.0 / –5.0 = –0.20 V  


(ii)
{from graph:} at 15 °C, R = 2.15 kΩ               (allow ±0.05 kΩ)
{Vout = - (Rf / Rin)  Vin}
reading = (2150 / 740) × 0.2
reading = 0.58 V (0.59 V → 0.57 V)  


(c)
(i)
{Since they are assumed to vary linearly, the voltmeter reading is inversely proportional to the temperature (the graph is a straight line with negative gradient).
29 °C corresponds to 0.35 V
1 °C corresponds to 0.35 × 29
15 °C corresponds to 0.35 × 29 / 15 = 0.68 V}
0.68 V

(ii) The resistance (of the thermistor) does not change linearly with temperature

Saturday, February 4, 2017

A straight conductor carrying a current I is at an angle θ to a uniform magnetic field of flux density B, as shown in Fig. 6.1.







Question 1
(a) A straight conductor carrying a current I is at an angle θ to a uniform magnetic field of flux density B, as shown in Fig. 6.1.


Fig. 6.1
The conductor and the magnetic field are both in the plane of the paper. State
(i) an expression for the force per unit length acting on the conductor due to the magnetic field, [1]

(ii) the direction of the force on the conductor. [1]

(b) A coil of wire consisting of two loops is suspended from a fixed point as shown in Fig. 6.2.


Fig. 6.2
Each loop of wire has diameter 9.4 cm and the separation of the loops is 0.75 cm.
The coil is connected into a circuit such that the lower end of the coil is free to move.
(i) Explain why, when a current is switched on in the coil, the separation of the loops of the coil decreases. [4]

(ii) Each loop of the coil may be considered as being a long straight wire.
In SI units, the magnetic flux density B at a distance x from a long straight wire carrying a current I is given by the expression
B = (2.0 × 10–7) I / x

When the current in the coil is switched on, a mass of 0.26 g is hung from the free end of the coil in order to return the loops of the coil to their original separation. Calculate the current in the coil. [4]





Reference: Past Exam Paper – November 2007 Paper 4 Q6





Solution 1:
(a)
(i) Force per unit length = BI sinθ

(ii) Direction: (downwards) into (the plane of) the paper
{Fleming’s left hand rule: Thumb = Force = ???, Forefinger = Field and Middle finger = Current = to right here.
Note that in this case, B and I are in same plane and the angle between them is NOT 90o. The force SHOULD be perpendicular to both B and I.}

(b)
(i)
{Direction of the magnetic field is obtained from the right hand grip rule}
EITHER
The magnetic field (due to the current) in one loop cuts / is normal to the current in the second, causing a force on the second loop.     EITHER Newton’s 3rd law discussed {From Newton’s 3rd law, a force will also act on the first loop and gives rise to an attraction between the 2 loops}                 OR vice versa clear gives rise to attraction

OR
Each loop acts as a coil which produces magnetic field. The fields are in the same direction {as the current in the 2 loops is in the same direction}, so the loops attract {causing the separation between them to decrease}.           

(ii)
F = BIL
Magnetic flux density, B = (2.0×10-7) I / (0.75×10-2) {= 2.67×10-5 I}
Force F = (Weight = mg =) (0.26×10-3) × 9.81 (= 2.55×10-3 N)        
{length of one loop, L = circumference = 2π r = 2π (d/2) = 2π (9.4×10-2) / 2 = 2π (4.7×10-2)}
2.55×10-3 = [(2.67×10-5) × I] × I × [2π (4.7×10-2)]    
Current I = 18 A
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