• Physics 9702 Doubts | Help Page 142

Question 700: [Electromagnetism]

A stiff straight copper wire XY is held fixed in uniform magnetic field of flux density 2.6 × 10⁻³ T, as shown in Fig.1.

Wire XY has length 4.7 cm and makes an angle of 34° with the magnetic field.

(a) Calculate force on the wire due to a constant current of 5.4 A in the wire.

(b) Current in the wire is now changed to an alternating current of r.m.s. value 1.7 A. Determine total variation in the force on the wire due to the alternating current.

Reference: Past Exam Paper – November 2014 Paper 41 & 42 Q6

Solution 700:

(a) Force F = BIL sinθ = (2.6 × 10^–3) × 5.4 × (4.7 × 10^–2) × sin 34° = 3.69 × 10^–4 N

(b)

Peak current = 1.7 × √2 = 2.4 A

Maximum force = (2.6 × 10^–3) × 2.4 × (4.7 × 10^–2) × sin 34° = 1.64 × 10^–4 N

{Force is a vector, it has both a magnitude and a direction. Since an a.c. is applied, the force will be maximum in one direction, then reduces to zero and then has a maximum value in the opposite direction. The magnitude of the force from the zero to the maximum value is as calculated above. But the total variation in the force is from the maximum value in one direction to the maximum value in the opposite direction. So, we take twice the value above.}

Total variation in force = 2 × (1.64 × 10^–4) = 3.3 × 10^–4 N

Question 701: [Work, Energy and Power]

Steel sphere is dropped vertically onto a horizontal metal plate. Sphere hits the plate with a speed u, leaves it at a speed v, and rebounds vertically to half of its original height.

Which expression gives the value of v / u ?

A 1 / 2²                                            B 1 / 2                         C 1 / √2                       D 1 – 1/√2

Reference: Past Exam Paper – June 2011 Paper 11 Q14

Solution 701:

Answer: C.

Let the distance fallen by the steel sphere be s.

For the fall,

Kinetic energy just before hitting the plate = Potential energy at maximum height

½ mu² = mgs

Speed u = √(2gs) = √2 (√gs)

For the sphere after rebound,

Kinetic energy just after hitting the plate = Potential energy at maximum height

½ mv² = mg (s/2)

Speed v = √(gs)

Ratio = v / u = 1 / √2

Question 702: [Direct Sensing]

Student designs an electronic sensor to monitor whether temperature in a refrigerator is above or below a particular value. The circuit is shown in Fig.1.

(a) Name components used in the output device.

(b) An operational amplifier (op-amp) is used as the processing unit. Describe function of this processing unit.

(c) State function of

(i) resistors C and D,

(ii) resistor B.

(d) Output device of the circuit in Fig.1 is changed so that the new output device may be used to switch on a high-voltage circuit.

(i) State component that is used in the new output device.

(ii) Draw on Fig.2 to show how component in (i), together with a diode, are connected so that the high voltage may be switched on when the output of the op-amp is negative.

Reference: Past Exam Paper – November 2012 Paper 43 Q9

Solution 702:

(a) Light-emitting diode

(b) The op-amp gives a high or a low output / +5 V or –5 V output that is dependent on which of the inputs is at a higher potential.

(c)

(i) This provides a reference/constant potential

(ii) It determines the temperature of the ‘switch-over’

(d)

(i) A relay

(ii) The relay should be connected correctly for op-amp output and high-voltage circuit and the diode with correct polarity in output from op-amp

{Current flows from a point of higher potential to a point of lower potential. Since the output of the op-amp is negative, current flows from earth to the op-amp. So, the flow is upwards. Thus, the diode should point upwards.}

Question 703: [Vectors]

(a) Explain the differences between the quantities distance and displacement.

(b) State Newton’s first law.

(c) Two tugs pull a tanker at constant velocity in direction XY, as represented in Fig.1.

Tug 1 pulls tanker with force T₁ at 25.0° to XY. Tug 2 pulls tanker with force of T₂ at 15.0° to XY. The resultant force R due to the two tugs is 25.0 × 10³ N in direction XY.

(i) By reference to the forces acting on the tanker, explain how tanker may be described as being in equilibrium

(ii)

1. Complete Fig.2 to draw a vector triangle for forces R, T₁ and T₂

2. Use vector triangle in Fig.2 to determine magnitude of T₁ and of T₂

Reference: Past Exam Paper – June 2012 Paper 23 Q1

Solution 703:

(a)

Displacement is a vector while distance is a scalar.

The displacement is the straight line between two points / distance is the sum of lengths moved / example showing difference

(b) Newton’s first law states that a body continues at rest or at constant velocity unless acted on by a resultant (external) force.

(c)

(i) The sum of T₁ and T₂ equals the frictional force. These two forces are in opposite directions.

(ii)

1. A scale vector triangle should be drawn with correct orientation / vector triangle with correct orientation both with arrows. The scale should be given or a mathematical analysis for the tensions

{Adding vectors graphically:

Adding vectors graphically is done by

1. firstly drawing the 1^st vector (can be any of the vectors present),

2. then drawing the next vector, with its tail at the head of the 1^st vector.

3. The resultant vector formed is a straight line which starts from the tail of the first vector and ends at the head of the last vector.

(So, the tail of the resultant vector starts at the tail of the 1^st vector drawn and its head is at the head of the last vector drawn)

Consider the diagram below as an example: }

{First, for T₂, lightly draw a line (long enough) at an angle of 15^o to the horizontal (at the right end – by using a protractor). This can be either above or below the horizontal resultant line.

Then for T₁, again lightly draw a line (long enough) at an angle of 25^o {from the rules of angles} to the horizontal (at the left end of the horizontal resultant line). This is drawn on the same side (either above or below the horizontal line) as T₁ was drawn.

The point of intersection indicates the end of the arrow for T₂ and the start of the arrow for T₁. Draw arrows on each of them and indicate clearly the directions of the vectors and their angles. An angle of about 140^o should be formed between T₁ and T₂.

Erase any unnecessary lines beyond the point of intersection.

[Alternatively, T₁ could be drawn first. …]

It can be measured that the horizontal resultant line representing the 25.0×10³N force is of length of about 12.3cm. So, 1.0cm represents about 2.0325N. The lengths of the vectors for T₂ and T₁ should be ABOUT 8.07cm and 4.97cm respectively. By using proportion, the answers of part (ii) can be obtained.}

2.

Magnitude of T₁ = 10.1 x 10³ (± 0.5 x 10³) N

Magnitude of T₂ = 16.4 x 10³ (± 0.5 x 10³) N

Question 704: [Current of Electricity > Charge]

Current in the circuit shown is 4.8 A.

What is the direction of flow and rate of flow of electrons through the resistor R?

direction of flow         rate of flow

A         X to Y                                     3.0 × 10¹⁹ s^–1

B         X to Y                                     6.0 × 10¹⁸ s^–1

C         Y to X                                     3.0 × 10¹⁹ s^–1

D         Y to X                                     6.0 × 10¹⁸ s^–1

Reference: Past Exam Paper – November 2010 Paper 11 Q31 & Paper 13 Q32

Solution 704:

Answer: C.

Electrons flow from the negative terminal of the cell (the shorter vertical arrow for the cell symbol). So, the direction of flow of electrons is from Y and X. [A and B are incorrect]

Charge Q = It

Charge Q = ne

where n is the number of electrons and e is the charge of an electron

ne = It

Rate of flow of electrons = n / t

Rate of flow of electrons = I / e = 4.8 / (1.6×10^-19) = 3.0 × 10¹⁹ s^–1