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Question 592: [Matter > Hooke’s law]

Two springs, one with spring constant k₁ = 4 kN m^–1 and other with spring constant k₂ = 2 kN m^–1, are connected as shown.

What is total extension of the springs when supporting a load of 80 N?

A 1.3 cm                     B 4 cm                         C 6 cm                         D 60 cm

Reference: Past Exam Paper – November 2014 Paper 13 Q25

Solution 592:

Answer: C.

This system consists of more than 1 spring attached in series to each other. The 2 strings each has its own spring constant. However, to calculate the extension, we need to find out the effective spring constant in the system.

For springs attached in series, the effective spring constant is given by

1 / k_eff = 1/k₁ + 1/k₂  

Spring constant, k_eff = [1/k₁ + 1/k₂]^-1 = [1/4 + 1/2]^-1 = 4/3 kN m^-1  

Spring constant, k_eff = 4/3 kN m^-1 = 4000/3 Nm^-1

Hooke’s law: F = ke

Extension e = F / k_eff = 80 / (4000/3) = 0.06m = 6cm

Question 593: [Electromagnetism]

(a) Define the tesla

(b) Two long straight vertical wires X and Y are separated by distance of 4.5 cm, as illustrated in Fig.1.

Wires pass through a horizontal card PQRS.

Current in wire X is 6.3 A in the upward direction. Initially, there is no current in wire Y.

(i) On Fig.1, sketch, in plane PQRS, the magnetic flux pattern due to the current in wire X. Show at least four flux lines.

(ii) Magnetic flux density B at a distance x from a long straight current-carrying wire is given by the expression

B = μ₀I / 2πx

where I is the current in the wire and μ₀ is the permeability of free space.

Calculate magnetic flux density at wire Y due to the current in wire X.

(iii) A current of 9.3 A is now switched on in wire Y. Use answer in (ii) to calculate the force per unit length on wire Y.

(c) Currents in the two wires in (b)(iii) are not equal.

Explain whether force per unit length on the two wires will be the same, or different.

Reference: Past Exam Paper – June 2013 Paper 42 Q5

Solution 593:

(a) The uniform magnetic flux normal to a long straight wire carrying a current of 1A creates a force per unit length of 1Nm^-1. (The tesla is a measure of the magnetic flux density.)

(b)

(i)

Sketch:

concentric circles

increasing separation (must show more than 3 circles)

correct direction (anticlockwise, looking down)

(right hand grip rule)

(ii)

B = (μ_oI)/(2πx) = (4π x 10^-7 x 6.3) / (2π x 4.5x10^-2) = 2.8x10^-5T

(iii) F = BILsin(θ)                               θ = 90^o

F/L = 2.8x10^-5 x 9.3 x 1 = 2.6x10^-4Nm^-1

(c) The force per unit length depends on the product of the two currents. OR From Newton’s third law, the action and reaction are equal and opposite. So, the force per unit length is the same for both.

Question 594: [Forces > Weight]

What is meant by weight of an object?

A the gravitational field acting on the object

B the gravitational force acting on the object

C the mass of the object multiplied by gravity

D the object’s mass multiplied by its acceleration

Reference: Past Exam Paper – June 2007 Paper 1 Q9

Solution 594:

Answer: B.

The weight of an object is the gravitational FORCE acting on the object. Weight is a force and its unit is newton. The gravitational field is a region in space where a mass experiences a force. [A is incorrect]

Usually, when we say gravity, we refer to the force of gravity. Weight is not the mass of the object multiplied by ‘gravity’, but ‘acceleration due to gravity’ or ‘gravitational field strength’. [C is incorrect]

A moving object (e.g. a car) may accelerate forwards, but this does not make it its weight. Weight is due to acceleration due to GRAVITY, not any other acceleration. [D is incorrect]

Question 595: [Current of Electricity]

A network of resistors, each of resistance R, is shown in Fig.1. 

(a) Calculate total resistance, in terms of R, between points

(i) A and C

(ii) B and X

(iii) A and Z

(b) Two cells of e.m.f. E₁ and E₂ and negligible internal resistance are connected into network in (a).

Currents in the network are as indicated in Fig.2. 

Use Kirchhoff’s laws to state the relation

(i) between currents I₁, I₂ and I₃

(ii) between E₂, R, I₂ and I₃ in loop BCXYB

(iii) between E₁, E₂, R, I₁ and I₂ in loop ABCXYZA

Reference: Past Exam Paper – June 2009 Paper 21 Q7

Solution 595:

(a)

(i) {Point B and C may be considered to be the same since there is no component connected between them.}

R

(ii) {B and Y are junctions at which current would change when going through the components connected to those junctions. So, the components between those junctions are in parallel.

[For components connected in series, the same current flows through them]

If another resistor is added in parallel across BY, then the total resistance between B and Y would change but the resistance between AB, across which a single resistor is present, remains the same (=R)}

{It should also be noted that points B and C can be considered to be the same point since there is no component between B and C. The same is true for points Y and X [that is the can be considered to be the same]. So, when considering the whole circuit, the resistance between BY, BX or CX are the same. But if we consider the resistor between B and Y or that between C and X separately, then the resistance between BY or CX are just R each.}

([1/R + 1/R]^-1 =) 0.5R

{Question: Why is it the case that the resistance between BY, BX and CX is the same? I can understand about BY and CX, but I don't understand about BX. In BX, why are we only considering the resistor on the line BCX? Why are we excluding the branch BY? I'm confused because: 1. The current branches out at the point B, so why exclude BY? 2. In other questions regarding voltage and resistance between two points, we usually include all branches in between.

Explanation: First of all, we are not excluding BY. Secondly, it is important to think of it as part of a complete circuit.  If we think of them separately, the resistance of BY is R, that between CX is R, resistance between BC is zero and resistance between XY is also zero. But this is not how we think of them. Think of it as part of a complete circuit, as will be explained below. Everything below will be about a complete circuit.

Consider the vertical wire between B and C. No component (resistor) is connected between B and C – so B and C can be considered to be the same point. The same goes for the vertical wire between X and Y.

Thus, the section BCXY can be drawn as follows:

It is then clear that the resistance BY, BX and CX are the same. Note, however, that this resistance between them is ‘0.5R’ (as calculate), and NOT ‘R’.} 

(iii) {The current flowing through the resistor between A and B is the same as that flowing through the resistor between Y and W. But this current is not the same flowing through the resistor between B and Y, nor the same as that flowing through C and X. So, between B and Y (as shown in the diagram), the resistors are in parallel.

So, the resistor between A and B is not in series with the resistor between B and Y, nor is it in series with the resistor between C and X. However, since the total current flowing the 2 resistors in parallel is the same as that flowing between AB and YZ, then the resistors between AB and YZ are in series with EQUIVALENT resistance between BY}

(R + [1/R + 1/R]^-1 + R =) 2.5R

(b)

(i) {From Kirchhoff’s law, the sum of currents entering a junction should be equal to the sum of currents leaving the junction. I₁ and I₂ are entering junction B and I₃ is leaving it.}

I₁ + I₂ = I₃

(ii) {We need to consider loop BCXYB. From Kirchhoff’s law, the sum of p.d.’s in a loop should be equal to the e.m.f. in the loop. The e.m.f. in the loop is E₂. The p.d. across B and Y is I₃R and the p.d. across the resistor between C and X is I₂R.}

E₂ = I₃R + I₂R

(iii) {We need to consider loop ABCXYZA. Again from Kirchhoff’s law, the sum of p.d.’s in a loop should be equal to the e.m.f. in the loop. The currents from E₁ and E₂ opposes each other. So, the overall e.m.f in the loop is obtained by subtracting the two (here we are assuming that E₁ is greater than E₂, that’s why we take E₁ – E₂). Between A and B, the p.d. is I₁R. Between the resistor in CX, the p.d. is I₂R. Across Y and Z, p.d. is I₁R (since current I₁ is flowing from the positive terminal of the battery E₁, the same current should return to its negative terminal). So, the sum of p.d. is I₁R + I₂R + I₁R.}

E₁ – E₂ = 2I₁R – I₂R

Question 596: [Electrostatic]

A small charge q is placed in the electric field of large charge Q.

Both charges experience a force F.

What is electric field strength of the charge Q at the position of the charge q?

A F / Qq                      B F / Q                                    C FqQ                         D F / q

Reference: Past Exam Paper – November 2009 Paper 11 Q27 & Paper 12 Q26 & November 2013 Paper 13 Q31

Solution 596:

Answer: D.

A small charge q is placed in the electric field of a large charge Q. Both charges experience force F.

Electric force F at position of charge q = Eq

Note that F also account for the magnitudes of the charges. The electric force can also be given by another formula (Coulomb’s law: F = Qq / 4πϵ₀r²) but we don’t need to calculate this here. The electric force is already given to be F.

Electric field is the force per unit charge. The charge in the expression is that on which the force acts (not the charge creating the field), so q is involved rather than Q.

Electric field strength of charge Q at position of charge q, E = F/q