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Question 1072: [Electromagnetism > Force between current-carrying wires]

Two long straight parallel copper wires A and B are clamped vertically. The wires pass through holes in a horizontal sheet of card PQRS, as shown in Fig. 7.1.

(a) There is a current in wire A in the direction shown on Fig. 7.1.

On Fig. 7.1, draw four field lines in the plane PQRS to represent the magnetic field due to the current in wire A.

(b) A direct current is now passed through wire B in the same direction as that in wire A. The current in wire B is larger than the current in wire A.

(i) On Fig. 7.1, draw an arrow in the plane PQRS to show the direction of the force on wire B due to the magnetic field produced by the current in wire A.

(ii) Wire A also experiences a force. State and explain which wire, if any, will experience the larger force.

(c) The direct currents in wires A and B are now replaced by sinusoidal alternating currents of equal peak values. The currents are in phase.

Describe the variation, if any, of the force experienced by wire B.

Reference: Past Exam Paper – June 2012 Paper 41 & 43 Q7

Solution 1072:

(a)

sketch: concentric circles (minimum of 3 circles)

separation increasing with distance from wire

correct direction

{The direction is obtained from the right hand grip rule.}

(b)

(i) arrow direction is from wire B towards wire A

{The direction of the force on wire B is obtained from Fleming’s left hand rule. Thumb = direction of force, forefinger = direction of magnetic field at wire B (this is towards SR, parallel to RQ) and middle finger = direction of current flowing in wire B (upwards).}

(ii)

EITHER reference to Newton’s third law

OR the force on each wire is proportional to the product of the two currents,

so the forces are equal

(c) The force is always towards wire A/ always in the same direction. It varies from zero (to a maximum value) {when the currents in both wires are zero}. The variation is sinusoidal / sin² (at) twice frequency of current {since the force depends on the product of the 2 currents (which are sinusoidal and in phase)}

Question 1073: [Current of Electricity]

Three identical cells each having an e.m.f. of 1.5 V and a constant internal resistance of 2.0 Ω are connected in series with a 4.0 Ω resistance R, firstly as in circuit (i), and secondly as in circuit (ii).

What is the ratio power in R in circuit (i) / power in R in circuit (ii)?

A 9.0               B 7.2               C 5.4               D 3.0               E 1.8

Reference: Past Exam Paper – J85 / I / 17

Solution 1073:

Answer: A.

Notice that in circuit (ii), the negative terminals of 2 of the cells are connected to each other, instead of the negative terminal being connected to the positive terminal of the other cell (the same goes for the positive terminal of that middle cell). Current flows in a specific direction, so this affects the overall e.m.f. in the circuit.

Overall e.m.f. in circuit (i) = 1.5 + 1.5 + 1.5 = 4.5 V

Overall e.m.f. in circuit (ii) = 1.5 – 1.5 + 1.5 = 1.5 V

In each case, the corresponding p.d. would be across the resistor R.

The total internal resistance due to the cells are the same in both cases. Internal resistance is due to the cells, it does not have a specific direction, so the total is the same in both cases.

Total internal resistance due to the cells = 2.0 + 2.0 + 2.0 = 6.0 Ω

Total resistance in each circuit = 6.0 + 4.0 = 10.0 Ω (same for both cases)

Power in resistor R = I²R = V² / R

Power P₁ in resistor R in circuit (i) = 4.5² / 4.0 = 5.0625 W

Power P₂ in resistor R in circuit (ii) = 1.5² / 4.0 = 0.5625 W

Ratio = P₁ / P₂ = 5.0625 / 0.5625 = 9.0

Question 1074: [Kinematics > Projectile motion]

An object is released from an airplane which is diving at an angle of 30^o from the horizontal with a speed of 50 ms^-1. If the plane is at a height of 4000 m from the ground when the object is released, find

(a) the velocity of the object when it hits the ground,

(b) the time taken for the object to reach the ground.

(Assume that the acceleration for the object due to gravity, g = 9.81 ms^-2)

Reference: Pacific Physics A Level Volume 1, by POH LIONG YONG, pg 82 Self Evaluation Exercise 5.3 Question 9

Solution 1074:

When released, the object will have the same speed as the airplane (that is, 50 ms^-1 at an angle of 30^o to the horizontal). The object is also initially at the height of 4000 m.

Acceleration due to gravity, g = 9.81 ms^-2

(a)

Consider the vertical (downward) component of the initial velocity: 50 sin(30)

{The vertical component of velocity is downward since the airplane is diving.}

Define the vertically downward direction as the positive direction.

Distance travelled by object = 4000 m

Let the vertical component of the velocity of the object when it hits the ground = v_v

v_v² = u² + 2as

v_v = √[{50sin30}² + 2(9.81)(4000)] = 281.3 ms^-1

Since the horizontal component (v_h) of the speed is not affected by any acceleration (or force), the magnitude of the velocity when it hits the ground is

Velocity² = v_v² + v_h²                (v_h = 50cos30 = 43.3 ms^-1)

Magnitude of velocity = √[281.3² + {50cos30}²] = 284.57 = 285 ms^-1

cos θ = 43.3 / 281.3

Angle θ (from horizontal) = cos^-1 (43.3 / 281.3) = 81.15°

Velocity is 285 ms^-1 at an angle of 81.15° from the horizontal.

(b)

Consider the vertical motion only since the object needs to travel a distance equal to the vertical height of 4000 m to reach the ground. The horizontal motion does not affect the time for the object to reach the ground.

Acceleration, a = (v – u) / t

Initial vertical speed, u = 50 sin(30) ms^-1

Final vertical speed, v = 281.3 ms^-1

9.81 = (281.3 – 50sin(30)) / t

Time taken, t = (281.3 – 50sin(30)) / 9.81 = 26.13 s