Chapter 4: Magnetic Field

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Chapter 4: Magnetic Field 4.1 Magnetic Field 4.1.1 Define magnetic field Magnetic field is defined as the region around a magnet where a magnetic force can be experienced.

Like poles (N-N or S-S) repel each other. Opposite poles (N-S) attract each other. 4.1.

The characteristics of magnetic field lines: The lines do not intersect one another The lines form a closed loop: magnetic field lines leave the North-pole and enter the South-pole.

1 Chapter 4: Magnetic Field 4.1 Magnetic Field Define magnetic field Magnetic field is defined as the region around a magnet where a magnetic force can be experienced. Magnetic field has two poles, called north (N) and south (S). These magnetic poles are always found in pairs whereas a single magnetic pole has never been found. Like poles (N-N or S-S) repel each other. Opposite poles (N-S) attract each other Identify magnetic field sources and sketch their magnetic field lines Magnetic field lines are used to represent a magnetic field. The characteristics of magnetic field lines: The lines do not intersect one another The lines form a closed loop: magnetic field lines leave the North-pole and enter the South-pole. The lines are closer together at the poles. (The number of lines per unit crosssectional area is proportional to the magnitude of the magnetic field.) Two sets of magnetic field lines can be superimposed to form a resultant magnetic field line. Magnetic field can be represented by crosses or by dotted circles as shown in figures below: Magnetic field lines enter the page perpendicularly Magnetic field lines leave the page perpendicularly BP3 1 FYSL

The pattern of the magnetic field lines can be

Bar magnet One bar magnet Horseshoe or U magnet

2 The pattern of the magnetic field lines can be determined by using two methods: Using compass needles Using sprinkling iron filings on paper Magnetic field sources: i. Bar magnet One bar magnet Horseshoe or U magnet Two bar magnets (unlike pole) Two bar magnets (like pole) BP3 2 FYSL

ii. Current-carrying conductor A stationary electric charge is surrounded by an electric field only. When an electric charge moves, it is surrounded by an electric field and a magnetic field.

3 ii. Current-carrying conductor A stationary electric charge is surrounded by an electric field only. When an electric charge moves, it is surrounded by an electric field and a magnetic field. The motion of the electric charge produces the magnetic field. A circular coil A long straightwire A solenoid iii. Earth magnetic field Note that the Earth s North Pole is really a south magnetic pole, as the north ends of magnets are attracted to it. Example N S N S N S S N BP3 3 FYSL

4.2 Magnetic Field Produced by Current-Carrying Conductor 4.2.1 Use magnetic field equations a) Long Straight Wire B r Magnitude of the magnetic field at any point from the conductor (wire): where B

4 4.2 Magnetic Field Produced by Current-Carrying Conductor Use magnetic field equations a) Long Straight Wire B r Magnitude of the magnetic field at any point from the conductor (wire): where B = magnetic field strength / magnetic flux density (T) I r 0I B 2r = current in the wire (A) = perpendicularly distance of B from the wire (m) μ o = constant of proportionality known as the permeability of free space (vacuum) = 4π x 10-7 Henry per metre (H m -1 ) View (a) from the top: The direction of magnetic field around the wire or coil can be determined by using the right hand grip rule as shown in Figure(b). Thumb direction of current Other fingers direction of magnetic field (clockwise OR anticlockwise) BP3 4 FYSL

b) Circular Coil Magnitude of the magnetic

ONE circular coil For N circular coils View

coil. µ 0 = permeability of free space 4π 10-7

The direction of magnetic field around the

right hand grip rule as shown in Figure(a).

5 b) Circular Coil Magnitude of the magnetic field at the centre of the circular coil: For ONE circular coil For N circular coils View from the top: where R = radius of the circular coil. µ 0 = permeability of free space 4π 10-7 H m -1 I = current N = number of coils (loops) The direction of magnetic field around the wire or coil can be determined by using the right hand grip rule as shown in Figure(a). B Example of multiple circular loops: (a) B (b) BP3 5 FYSL

c) Solenoid The magnitude of magnetic field intensity at the

from the top: Since, therefore it can be written as The

solenoid is given by: where n = numbers of turns per unit

current N = number of coils (loops) The directions of the

solenoid from both end or applying the right hand grip rule as

6 c) Solenoid The magnitude of magnetic field intensity at the centre (mid-point/ inside) of N turn solenoid is given by View from the top: Since, therefore it can be written as The magnitude of magnetic field intensity at the end of N turn solenoid is given by: where n = numbers of turns per unit length µ 0 = permeability of free space 4π 10-7 H m -1 I = current N = number of coils (loops) The directions of the fields can be found by viewing the current flows in the solenoid from both end or applying the right hand grip rule as shown in Figure (a). (a) Thumb north pole Other fingers direction of current in solenoid. BP3 6 FYSL

7 Example Determine the magnetic field strength at point X and Y from a long, straight wire carrying a current of 5 A as shown below. Solution A circular coil having 400 turns of wire in air has a radius of 6 cm and is in the plane of the paper. What is the value of current must exist in the coil to produce a flux density of 2 mt at its center? An air-core solenoid with 2000 loops is 60 cm long and has a diameter of 2.0 cm. If a current of 5.0 A is sent through it, what will be the flux density within it? A solenoid is constructed by winding 400 turns of wire on a 20 cm iron core. The relative permeability of the iron is What current is required to produce a magnetic induction of 0.5 T in the center of the solenoid? BP3 7 FYSL

in the opposite direction. Solution Exercise Two long straight wires are oriented perpendicular to the page as shown in Figure below. The current in one wire is I 1 = 3.

8 Two straight parallel wires are 30 cm apart and each carries a current of 20 A. Find the magnitude and direction of the magnetic field at a point in the plane of the wires that is 10 cm from one wire and 20 cm from the other if the currents are i. in the same direction, ii. in the opposite direction. Solution Exercise Two long straight wires are oriented perpendicular to the page as shown in Figure below. The current in one wire is I 1 = 3.0 A pointing into the page and the current in the other wire is I 2 = 4.0 A pointing out of the page. Determine the magnitude and direction of the nett magnetic field intensity at point P. (Given 0 =4 x 10-7 H m -1 ) Answer: T, 63.1 below +ve x-axis A 2000 turns solenoid of length 40 cm and resistance 16 is connected to a 20 V supply. Find the magnetic flux density at the end of the axis of the solenoid. Answer: T BP3 8 FYSL

4.3 Force on a Moving Charged Particle in a Uniform Magnetic Field 4.3.1 Use magnetic force equation A stationary electric charge in a magnetic field will not experience any force.

9 4.3 Force on a Moving Charged Particle in a Uniform Magnetic Field Use magnetic force equation A stationary electric charge in a magnetic field will not experience any force. But if the charge is moving with a velocity, v in a magnetic field, B then it will experience a force. This force known as magnetic force. The magnitude of the magnetic force can be calculated by using the equation below : F q v B F qvbsin where q : magnitude of the charge : angle between and The direction of the magnetic force can be determined by using right hand rule: F IMPORTANT! v B Thumb indicates the direction of the magnetic force exerted on a positive charge. If the charge is negative, direction of the force is opposite. Example Determine the direction of the magnetic force, exerted on a charge in each problems below: BP3 9 FYSL

10 4.3.2 Describe circular motion of a charge in a uniform magnetic field Use relationship F B = F C Consider a charged particle moving in a uniform magnetic field with its velocity perpendicular to the magnetic field. As the particle enters the region, it will experience a magnetic force which the force is perpendicular to the velocity of the particle. Hence the direction of its velocity changes but the magnetic force remains perpendicular to the velocity. Since the path is circle therefore the magnetic force F B contributes the centripetal force F c (net force) in this motion. Thus FB F C 2 mv Bqvsin and θ = 90 r mv r Bq where m : mass of the charged particle v : magnitude of the velocity r : radius of the circular path q : magnitude of the charged particle The period of the circular motion, T makes by the particle is given by T 2r v T 2m Bq The frequency of the circular motion makes by the particle is given by f 1 T f Bq 2m BP3 10 FYSL

Example A charge q 1 = 25.0 μc moves with a speed of 4.5 x 10 3 m s -1 perpendicularly to a uniform magnetic field. The charge experiences a magnetic force of 7.31 x 10-3 N. A second charge q 2 = 5.

11 Example A charge q 1 = 25.0 μc moves with a speed of 4.5 x 10 3 m s -1 perpendicularly to a uniform magnetic field. The charge experiences a magnetic force of 7.31 x 10-3 N. A second charge q 2 = 5.00 μc travels at an angle of 40.0 o with respect to the same magnetic field and experiences a 1.90 x 10-3 N force. Determine a. The magnitude of the magnetic field b. The speed of q 2. Solution An electron at point A in figure above has a speed v of m s -1. Determine the magnitude and direction of the magnetic field that will cause the electron to follow the semicircular path from A to B. (Given e = C and m e = kg) Exercise Calculate the magnitude of the force on a proton travelling 3.0 x10 7 m s -1 in the uniform magnetic flux density of 1.5 Wb m -2, if a. the velocity of the proton is perpendicular to the magnetic field. b. the velocity of the proton makes an angle 50 with the magnetic field. (Given the charge of the proton is 1.60 x C) Answer: N, N An electron is moving in a magnetic field. At a particular instant, the speed of the electron is 3.0 x 10 6 m s -1. The magnitude of the magnetic force on the electron is 5.0 x N and the angle between the velocity of the electron and the magnetic force is 30. Calculate the magnitude of the magnetic field. Given e = C. Answer: 1.2 T A proton is moving with velocity 3 x 10 5 m s -1 vertically across a magnetic field 0.02 T. Calculate a. kinetic energy of the proton b. the magnetic force exerted on the proton c. the radius of the circular path of the proton. (m p = 1.67 x kg) Answer: 7.52 x J, 9.6 x N, 0.16 m BP3 11 FYSL

4.4 Force on a Current Carrying Conductor in a Uniform Magnetic Field 4.4.1 Use magnetic force equation When a current-carrying conductor is placed in a magnetic field B, thus a magnetic force will acts on that conductor.

12 4.4 Force on a Current Carrying Conductor in a Uniform Magnetic Field Use magnetic force equation When a current-carrying conductor is placed in a magnetic field B, thus a magnetic force will acts on that conductor. The magnitude of the magnetic force exerts on the current-carrying conductor is given by F I l B F IlBsin where l : length of the conductor : angle between l and The direction of the magnetic force can be determined by using right hand rule: F l B One tesla is defined as the magnetic flux density of a field in which a force of 1 newton acts on a 1 metre length of a conductor which carrying a current of 1 ampere and is perpendicular to the field. Example Determine the direction of the magnetic force, exerted on a conductor carrying current, I in each problems below. BP3 12 FYSL

13 Example A wire of 20 cm long is placed perpendicular to the magnetic field of 0.40 Wb m -2. a. Calculate the magnitude of the force on the wire when a current 12 A is flowing. b. For the same current in (a), determine the magnitude of the force on the wire when its length is extended to 30 cm. c. If the force on the 20 cm wire above is 60 x 10-2 N and the current flows is12 A, find the magnitude of magnetic field was supplied. Solution A square coil of wire containing a single turn is placed in a uniform 0.25 T magnetic field. Each side has a length of 0.32 m, and the current in the coil is 12 A. Determine the magnitude of the magnetic force on each of the four sides. Exercise A straight wire with a length of 0.65 m and mass of 75 g is placed in a uniform magnetic field of 1.62 T. If the current flowing in the wire is perpendicular to the magnetic field, calculate the current required to balance the wire? (g = 9.81 ms -2 ) Answer: 0.70 A A wire of length m carries a current of 21.0 A. In the presence of a T magnetic field, the wire experiences a force of 5.46 N. What is the angle (less than 90 o ) between the wire and the magnetic field? Answer: BP3 13 FYSL

4.5 Forces between Two Parallel Current-Carrying Conductors 4.5.1 Derive magnetic force per unit length of two parallel current-carrying conductors 4.

currents I 1 and I 2 with length l are placed parallel to each other as shown in figure below.

View from the top Force exerted on conductor X Force exerted on conductor Y The magnitude of the magnetic flux

the page Conductor X carries a current I 1 and in the magnetic field B 2 then conductor X will experiences a

The magnitude of the magnetic flux density, B 1 at point P on conductor Y due to the current in conductor X is

conductor Y will experiences a magnetic force, F 12.

14 4.5 Forces between Two Parallel Current-Carrying Conductors Derive magnetic force per unit length of two parallel current-carrying conductors Use magnetic force per unit length equation Consider two identical straight conductors X and Y carrying currents I 1 and I 2 with length l are placed parallel to each other as shown in figure below. The conductors are in vacuum and their separation is d. View from the top Force exerted on conductor X Force exerted on conductor Y The magnitude of the magnetic flux density, B 2 at point Q on conductor X due to the current in conductor Y is given by: B 2 0I 2 2d Direction: out of the page Conductor X carries a current I 1 and in the magnetic field B 2 then conductor X will experiences a magnetic force, F 21. The magnitude of the magnetic flux density, B 1 at point P on conductor Y due to the current in conductor X is given by B 1 0I1 2d Direction: into the page Conductor Y carries a current I 2 and in the magnetic field B 1 then conductor Y will experiences a magnetic force, F 12. The magnitude of F 21 is given by F F 21 B2I1l sin 0I 2 I 2d 21 1l 0I 2 I l 2d F21 1 and sin90 F 12 F 21 F 90 Direction: towards conductor Y 0I1I2L 2d The magnitude of F 12 is given by F 12 B1 I2l sin 0I1 I 2d 12 2l BP3 14 FYSL F 0I1 F12 I 2l 2d and sin90 90 Direction: towards conductor X Conclusion: Type of the force is attractive force

If the direction of current in conductor Y is change to upside down as shown in figure View from the top The magnitude of F 12 and F 21 can be determined by using equations above and its direction by

0 A currents hence the force per unit length that each conductor exerts on the other is F L F L 0I1I 2d 2 2.0x10 7 N m 1 F L 7 ( 4πx10 )( 10. )( 10. ) 2π( 10.

15 If the direction of current in conductor Y is change to upside down as shown in figure View from the top The magnitude of F 12 and F 21 can be determined by using equations above and its direction by applying right hand rule. Conclusion: Type of the force is repulsive force Define one ampere If two long, straight, parallel conductors, 1.0 m apart in vacuum carry equal 1.0 A currents hence the force per unit length that each conductor exerts on the other is F L F L 0I1I 2d 2 2.0x10 7 N m 1 F L 7 ( 4πx10 )( 10. )( 10. ) 2π( 10. ) The ampere (one ampere) is defined as the current that flowing in each of two parallel conductors which are 1.0 metre apart in vacuum, would produce a force per unit length between the conductors of 2.0 x 10-7 N m -1. Example Two long straight parallel wires are placed 0.25 m apart in a vacuum. Each wire carries a current of 2.4 A in the same direction. a. Sketch a labelled diagram to show clearly the direction of the force on each wire. b. Calculate the force per unit length between the wires. c. If the current in one of the wires is reduced to 0.64 A, calculate the current needed in the second wire to maintain the same force per unit length between the wires as in (b). (Given µ 0 = 4π 10-7 T m A -1 ) Solution BP3 15 FYSL

4.6 Torque on a Coil 4.6.1 Use torque equation Torque is the tendency

Top view Side view For a coil of N turns, the magnitude of the torque

the coil B : magnetic flux density I : current flows in the coil θ :

16 4.6 Torque on a Coil Use torque equation Torque is the tendency of a force to rotate an object about an axis. Top view Side view For a coil of N turns, the magnitude of the torque is given by NI A B NIAB sin where τ : torque on the coil A : area of the coil B : magnetic flux density I : current flows in the coil θ : angle between vector area, A (the normal to plane of the coil) and B N : number of turns (coils) BP3 16 FYSL

4.6.2 Explain the working principles of a moving coil galvanometer Structure of a moving-coil galvanometer The galvanometer is the main component in analog meters for measuring current and voltage.

17 4.6.2 Explain the working principles of a moving coil galvanometer Structure of a moving-coil galvanometer The galvanometer is the main component in analog meters for measuring current and voltage. It consists of a magnet, a coil of wire, a spring, a pointer and a calibrated scale. The coil of wire contains many turns and is wrapped around a soft iron cylinder. The coil is pivoted in a radial magnetic field, so that no matter what position it turns, the plane of the coil is always parallel to the magnetic field. The basic operation of the galvanometer uses the fact that a torque acts on a current loop in the presence of a magnetic field. When there is a current in the coil, the coil rotates in response to the torque (τ = NIAB ) applied by the magnet. This causes the pointer (attached to the coil) to move in relation to the scale. The torque experienced by the coil is proportional to the current in it; the larger the current, the greater the torque and the more the coil rotates before the spring tightens enough to stop the rotation. Hence, the deflection of the pointer attached to the coil is proportional to the current. The coil stops rotating when this torque is balanced by the restoring torque of the spring. Torque due to magnetic force restoringtorque NIAB k k I NAB where k is the stiffnessconstant of the spring (torsionalconstant) θ : rotation angle of the coil in radian From this equation, the current I can be calculated by measuring the angle θ. This working principles of a moving coil galvanometer also used in voltmeter (multiplier), ammeter (shunt), ohmmeter and multimeter. s BP3 17 FYSL

18 Example A 20 turns rectangular coil with sides 6.0 cm x 4.0 cm is placed vertically in a uniform horizontal magnetic field of magnitude 1.0 T. If the current flows in the coil is 5.0 A, determine the torque acting on the coil when the plane of the coil is a. perpendicular to the field, b. parallel to the field, c. at 60 to the field. Solution Exercise A rectangular coil of 10 cm x 4.0 cm in a galvanometer has 50 turns and a magnetic flux density of 5.0 x 10-2 T. The resistance of the coil is 40 and a potential difference of 12 V is applied across the galvanometer, calculate the maximum torque on the coil. Answer: 3.0 x 10-3 N m. A moving coil meter has a 50 turns coil measuring 1.0 cm by 2.0 cm. It is held in a radial magnetic field of flux density 0.15 T and its suspension has a torsional constant of N m rad -1. Determine the current is required to give a deflection of 0.5 rad. Answer: A Calculate the magnetic flux density required to give a coil of 100 turns a torque of 0.5 Nm when its plane is parallel to the field. The dimension of each turn is 84 cm 2, and the current is 9.0 A. Answer: 66.1 mt BP3 18 FYSL

19 4.7 Motion of Charged Particle in Magnetic Field and Electric Field Explain the motion of a charged particle in both magnetic field and electric field Derive and use velocity equation in a velocity selector Velocity Selector Velocity selector consists of a pair of parallel metal plates across which an electric field can be applied in uniform magnetic field. Consider a positive charged particle with mass m, charge q and speed v enters a region of space where the electric and magnetic fields are perpendicular to the particle s velocity and to each other as shown in figure above. The charged particle will experiences the electric force F E is downwards with magnitude qe and the magnetic force F B is upwards with magnitude Bqv as shown in figure above. If the particle travels in a straight line with constant velocity hence the electric and magnetic forces are equal in magnitude (dynamic equilibrium). Therefore FB F E qvbsin90 qe E v B Only particles with velocity equal to E/B can pass through without being deflected by the fields. Equation above also works for electron or other negative charged particles. Mass Spectrometer When the charged particle entering the region consists of magnetic field only, the particle will make a semicircular path of radius r. FB F C 2 mv Bqv and r q E 2 m rb E v B BP3 19 FYSL

20 Example A velocity selector is to be constructed to select ions (positive) moving to the right at 6.0 km s -1. The electric field is 300 Vm -1 upwards. What should be the magnitude and direction of the magnetic field? Solution An ion enters a uniform magnetic field B. The path of the ion is a spiral as shown in the figure. a) Determine whether the ion is positively or negatively charged. b) Give reasons for the shape of the path. Exercise An electron moving at a steady speed of m s 1 passes between two flat, parallel metal plates 2.0 cm apart with a potential difference of 100 V between them. The electron is kept travelling in a straight line perpendicular to the electric field between the plates by applying a magnetic field perpendicular to the electron s path and to the electric field. Calculate : a. the intensity of the electric field. b. the magnetic flux density needed. Answer : V m 1 ; T An electron with kinetic energy of J passes perpendicular through a uniform magnetic field of T. It is found to follow a circular path. Calculate a. the radius of the circular path. b. the time required for the electron to complete one revolution. (Given e = C kg -1, m e = kg) Answer: m, s BP3 20 FYSL

Physics Week 5(Sem. 2) Name. Magnetism. Chapter Summary. Magnetic Fields

Physics Week 5(Sem. 2) Name. Magnetism. Chapter Summary. Magnetic Fields Physics Week 5(Sem. 2) Name Chapter Summary Magnetism Magnetic Fields Permanent magnets have long been used in navigational compasses. The needle in a compass is supported to allow it to freely rotate