PHY 102 Lecture 6 6 1 Magnetic Field

PHY 102: Lecture 6 • • 6. 1 Magnetic Field 6. 2 Magnetic Force on Moving Charges 6. 3 Magnetic Force on Currents 6. 4 Magnetic Field Produced by Current

PHY 102: Lecture 6 Magnetic Force Magnetic Fields 6. 1 Magnetic Field

Basics of This Chapter • Charge moving in a magnetic field feels a force • Moving charge creates a magnetic field

Magnetic Field Into and Out of Page

Magnetic Poles • North magnetic pole – The end of the needle that points north • South magnetic pole – The opposite end • No one has been able to separate magnetic poles • Electric charges can be separated

Magnetic Force • Direction of the magnetic force – Like poles repel each other – Unlike poles attract each other

Magnetic Field • A magnetic field is created at all points in space surrounding a magnet • The magnetic field at each point is a vector. It has both a magnitude called the magnetic field strength B, and a direction • The magnetic field exerts forces on magnetic poles. The force on a north magnetic pole is parallel to B; the force on a south magnetic pole is opposite to B • Magnetic forces cause a compass needle to become aligned parallel to a magnetic field, with the north pole of the compass showing the direction of the magnetic field at that point

Magnetic Field Lines • Magnetic field lines – To visualize the electric field, we introduced electric field lines – It is possible to draw magnetic field lines – The lines originate from north pole, and end on the south pole – They do not end in midspace – The strength of the field is proportional to the number of lines per unit area that passes through a surface oriented perpendicular to the lines

Magnetic Pole Force Directions - 1

Magnetic Pole Force Directions - 2 • At any location in the vicinity of a magnet, the north pole of a small compass needle points in the direction of the magnetic field at that location

Magnetic Field Lines

Earth’s North Magnetic Pole • The compass and field lines point to the earth’s north magnetic pole • Therefore, the earth’s north magnetic pole is actually a south pole

PHY 102: Lecture 6 Magnetic Force Magnetic Fields 6. 2 Magnetic Force on Moving Charge

Magnetic Field Force on Moving Charge • Two conditions must be met for a charge to experience a magnetic force when placed in a magnetic field – The charge must be moving – The velocity of the moving charge must have a component that is perpendicular to the direction of the magnetic field

No Force – Velocity Parallel to Magnetic Field

Max Force – Velocity Perpendicular to Magnetic Field

Some Force – Velocity at Angle to Magnetic Field

Direction of Magnetic Force Right-Hand Rule No. 1 • Extend the right hand so the fingers point along the direction of the magnetic field B • The thumb points along the velocity v of the charge • The palm of the hand then faces in the direction of the magnetic force F that acts on a positive charge • If the moving charge is negative, the direction of the magnetic force is opposite to that predicted by RHR-1

Illustration of Right-Hand Rule No. 1

Definition of Magnetic Field • The magnitude B of the magnetic field at any point in space is defined as • • F is the magnitude of magnetic force on a test charge |q 0| is the magnitude of the test charge v is the magnitude of the charge’s velocity v makes an angle (0 <= 1800) with the direction of the magnetic field • The magnetic field B is a vector, and its direction can be determined by using a small compass needle • SI Unit of Magnetic Field = 1 tesla (T)

Problem 1 - 1 • A proton in a particle accelerator has a speed of 5. 0 x 106 m/s • The proton encounters a magnetic field whose magnitude is 0. 40 T • Direction of the field makes and angle of q = 30. 00 with respect to the protons velocity • (a) Find the magnitude and direction of the magnetic force on the proton • (b) The acceleration of the proton

Problem 1 - 2 ü (a) Positive charge on proton is 1. 60 x 10 -19 C ü Magnitude of force is F = |q 0|v. B sinq ü The direction of the magnetic force is given by RHR-1 and is directed upward

Problem 1 - 3 ü (b) Acceleration if from Newton’s second law ü a = F/m = 1. 6 x 10 -13 / 9. 11 x 10 -31 = 1. 8 x 1017 m/s 2

Compare Particle Motion in Electric and Magnetic Field • Positive charge moving between the plates of a parallel plate capacitor • Initially, the charge is moving perpendicular to the electric field • Direction of the electric force is in the same direction as the electric field • Particle is deflected sideways

Compare Particle Motion in Electric and Magnetic Field • Initially, positive charge is moving at right angles to a magnetic field • When the charge enters the field it is deflected upwards by magnetic force • As charge moves upward, the direction of the magnetic force changes • Force always remains perpendicular to the magnetic field and velocity • Charge moves in circle

Velocity Selector • Force of the electric field on particle is q. E in downward direction • Force of the magnetic field on the moving particle is qv. B in the upward direction • The forces are balanced when • q. E = qv. B • v = E/B • This is the velocity of the particle that will pass straight through the velocity filter from one end to the other

Work Done on a Charged Particle Moving Through Magnetic Field • The force is perpendicular to the direction of motion • Therefore, the work done by the magnetic field is zero

Circular Trajectory - 1 • Velocity of positively charged particle is perpendicular to a constant magnetic field • The magnetic force moves the particle in a circular path • At point 1, the magnetic force F points upward • This force causes the parth to bend upward

Circular Trajectory - 2 • At point 2, the magnetic force is directed to the left • The magnetic force always remains perpendicular to the velocity • Force is directed toward the center of the circular path

Circular Trajectory - 3 • Centripetal force equals magnetic force

Problem 2 - 1 • Proton is released from rest at point A next to positive plate of parallel plate capacitor • Proton accelerates towards the negative plate • Proton leaves capacitor at point B through a small hole • Electric potential of positive plate is 2100 V greater than the negative plate ü VA – VB = 2100 V

Problem 2 - 2 • Outside the capacitor, the proton travels at a constant velocity until it enters a region of constant magnetic field • Field is directed out of the page ü B = 0. 10 T • Velocity is perpendicular to the magnetic field

Problem 2 - 3 • (a) Find the speed v. B of the proton when it leaves the negative plate of the capacitor • (b) Find the radius r of the circular path on which the proton moves in a magnetic field

Problem 2 - 4 • (a) • Electric force is conservative • Energy is conserved while in capacitor

Problem 2 - 5 • (b) The radius of curvature of the circular path in the magnetic field is

Problem 3 - 1 • Bubble-chamber tracks resulting from event at point A • A gamma ray traveling from the left spontaneously transforms into two charged particles • No track from gamma ray • Particles move away from point A producing two spiral tracks • A third charged particle is knocked out of a hydrogen atom and moves forward producing the long track with the slight upward curvature

Problem 3 - 2 • Each particle has the same mass and a charge of the same magnitude • A uniform magnetic field is directed out of the page • Based on RHR-1 and formula for radius of curved track in magnetic field, we know: ü Particles 1 and 3 are – ü Particle 2 is + ü Particles 1 and 2 are losing energy to hydrogen collisions because tracks spiral inward ü Particle 3 has greatest speed

Mass Spectrometer - 1

Mass Spectrometer - 2 • Conservation of energy in electric field • Radius of curvature in magnetic field

Mass Spectrometer Output

Dr. Molof Thesis Rb 85, 87 Special Spectrometer Scan

PHY 102: Lecture 6 Magnetic Force Magnetic Fields 6. 3 Magnetic Force on Currents

Force on Current in Magnetic Field

Problem 4 - 2 Rail Gun • • Two conducting rails are 1. 6 m apart Rails are parallel to the ground at the same height A 0. 20 -kg aluminum rod is lying on top of the rails A 0. 050 -T magnetic field points upward The field is perpendicular to the ground There is a current in the rod The coefficient of static between rod and rails is s = 0. 45

Problem 4 - 2 Rail Gun • How much current is needed to make the rod begin to move? • In what direction will it move?

Problem 4 - 3 Rail Gun • • Friction force = f = s. N = smg f = 0. 45 x 0. 20 x 9. 8 = 0. 882 N Magnetic force = ILB = I(1. 6)(0. 050) Magnetic force = (0. 08)I N 0. 882 = 0. 08 I I = 0. 882 / 0. 08 = 11. 0 A If current is positive RHR-1 give motion to left, negative to the right

PHY 102: Lecture 6 Magnetic Force Magnetic Fields 6. 4 Magnetic Field Produced by Current

Right Hand Rule No. 2 • Curl the fingers of the right hand into the shape of a half-circle • Point the thumb in the direction of the conventional current I • The tips of the fingers will point in the direction of the magnetic field B

Magnetic Field Long Straight Wire 0 = 4 x 10 -7 Tm/A This is permeability of free space r is distance from wire

Problem 5 - 1 • Long straight wire carries current of 3. 0 A • Particle charge of +6. 5 x 10 -6 C • Particle moving parallel to wire at distance of 0. 050 m • Particle speed is 280 m/s • Find magnitude and direction of the magnetic force exerted on the charged particle by the current in the wire.

Problem 5 - 2 • Magnetic field produced by wire • Force on charge F= • Direction magnetic force is radially inward toward wire (RHR-1)

Problem 6 - 1 • • Two long, parallel, straight wires Wires separated by 0. 065 m Current is I 1 = 15 A and I 2 = 7. 0 A Find the magnitude and direction of force that magnetic field of wire 1 applies to a 1. 5 m section of wire 2

Problem 6 - 2 • Magnetic force on part of wire 2 • Magnetic field produced by wire 1 • Combined

Problem 6 - 3 • Using RHR-1 the force on wire 2 is away from wire 1 • Using Newton’s Third Law, the force on wire 1 is away from wire 2 • If the current in wire 2 is reversed, the direction of the forces is reversed

Magnetic Field Loop of Wire • Field at center of loop • N is number of coils in loop • R is the radius of the loop

Loop of Wire acts like Bar Magnet • Current-carrying loop acts like a bar magnet

Two Loop of Wire acts like Repel/Attract Bar Magnets • When the currents are in the same direction the loops attract • When the currents are in opposite directions the loops repel • Used RHR-1

Magnetic Field Long Solenoid • Field inside of solenoid • N is number of coils in loop

Ampere’s Law • Constant magnetic field Dl is a small segment of length along a closed path of arbitrary shape round the current • B|| is the component of the magnetic field parallel to Dl, • I is the net current passing through the surface bounded by the path 0 is the permeability of free space • Symbol S indicates that the sum of all terms must be taken around the closed path

Problem 7 - 1 • Use Ampere’s law to obtain the magnetic field produced by the current in an infinitely long, straight wire • Along the circular path the magnetic field is parallel to Dl and has constant magnitude

Problem 7 - 2