Magnetism

Effects of magnetism have been known since antiquity: lodestone used to find north

Connection between magnetism and electricity was demonstrated in 1820 by Hans Christrian Oersted.

Electric field = E field, a vector Magnetic field = B field, a vector

Permanent magnets: [bar magnets, or horseshoe magnets, or round magnets ]

In electricity, + and – charges can exist separately.

All magnets are bipoles = have 2 poles + pole and – pole

Opposite poles attract, same poles repel

Breaking a magnet in half results in formation of 2 magnets

A “magnetic monopole” does not exist.

A magnet creates a magnetic field around it.

Field lines directed away from N pole, Directed towards S pole.1

The direction of the magnetic field at a given location is the direction in which the NORTH POLE of a compass points when placed at that location.

Magnetic field lines run from N pole, to S pole and continue through the magnet to complete a loop.

Mag field lines DO NOT CROSS

A refrigerator magnetic only sticks on one side. Refrigerator magnet is an array of tiny horseshoe magnets.

2

GEOMAGNETISM

The Earth produces its own Bfield The Earth behaves as a giant bar magnet

The poles of Earth’s Bfield is not perfectly aligned with Earth’s axis.Magnetic North varies slowly over time. Magnetic North is tilted

about 11.5 degrees from “true North”

Earth’s B field lines are approximately parallel to the Earth’s surface.

When magma [molten rock from beneath crust]extrudes and cools, the rock becomes magnetized and domains align with Earth’s magnetic field lines.

Over millions of years, the direction of N has reversed several times. The last reversal was 780,000 years ago.This is recorded in the rocks formed out of the Mid-Atlantic Ridge

3

MAGNETIC FORCE ON MOVING CHARGES

Consider a B field that points from left to right.

A particle of charge q moves with velocity v at angle wrt the Bfield

The magnitude of the magnetic force acting on the charged particle is

F = |q |v Bsin unit of force = Newton

Magnetic force depends on both the charge q of the particle and the magnitude of the Bfield.

Behavior of particles in Bfields differs from behavior in Efields:

4

A charged particle must be MOVING in Bfield in order to have force exerted on it by Bfield. [if v=0, F=0]

If a charged particle moves parallel to Bfield then no force exerted on it by Bfield.[if sin= 0, then F=0]

Maximum force exerted on charged particle if velocity is perpendicular to Bfield.

Sign of charge determines the direction of the magnetic force on the particle.

Magnitude of magnetic field B B = F____ |q|v sin

Unit of Bfield = tesla, 1T = 1 N A-m

Problem 1: Particle 1, q1= 3.60 C and v1= 862 m/sec travels at right angles to a uniform Bfield. The magnetic force q1 experiences is 4.25X10-

3NParticle 2, q2=53.0mC and v2=1300 m/sec moves at angle 55degrees wrt Bfield. Find a) the strength of the Bfield, and (b) the magnitude of the magnetic force exerted on q2.

The tesla is a large unit of magnetic strength.

The magnetic field on the surface of Earth is 5T.

Smaller unit is the gauss , G

1 gauss = 10-4 Tesla 1G = 10-4 T

5

Magnetic force right hand rule:

Direction of magnetic force F of a moving positive charge points in direction that is perpendicular to both velocity v and Bfield given by right hand rule

F = qv X B

If directions are given by

Cross product: X xY = Z Y x Z =X Z x X= y

Y x X = - Z Z x Y = -X X x Z = - Y

A magnetic field perpendicular and into page

is shown as “feathers of an arrow”

6

A Bfield perpendicular to page and out of page is shown as “tip of an arrow”

So the force exerted on charged positive and negative particles on Bfield

Problem 2: three particles travel through a magnetic field as shown. A)For each particle, state whether the charge of the particle is +, - or 0.

B) for each, sketch the direction of the force acting on the particle.

Problem 3. A particle of charge 7.70mC and speed 435 m/sec is acted on by both electric and magnetic fields. The particle moves along the 7

+xdirection along the x axis. The Bfield = 3.20 T and points in +ydirection and Efield = 8100 N/C points in +Zdirection. Determine the magnitude and direction of force acting on the particle.

Electric force F = qE

magnetic force F = qvBsin

motion of charged particle in Efield: motion of charged particle in B field

Particle behaves like mass in gravity field. Particle travels in circular path.

An electric field can do work on a charged particle [in example above component of the velocity will be in the direction of Efield as soon as particle moves downward. ]

A constant magnetic field CANNOT do work on a charged particle because the Bfield is always perpendicular to the direction of motion.

Velocity selector: charged particles move through region of space that has both Efield and Bfield. If the speed of the particle has a particular value, v, the net force acting on the particle is zero and the particle will not be deflected. Any charged particle not having speed v will be deflected.

8

For velocity selector: v = E/B

If a charged particle travels // to a Bfield, it travels in a constant straight line.

Circular motion: if velocity of charged particle of mass m and charge q is perpendicular to a Bfield, the particle will move in a circle of radius r.

The magnetic force will be the centripetal force that keeps the particle moving in a circle.

Fc = mac = mv2/r

Fc = |q|vB So qvB = mv2/r Radius r = mv/qB

A mass spectrometer is used to separate charged ions [same atomic number, different atomic mass] Since the radius depends on mass, each ion will have a different radius

Problem 4: two singly ionized uranium isotopes, U-235 and U-238 are sent into a mass spec at velocity 1.05e5 m/sec.Mass U-235 = 3.90e-25 kg Mass U-238 = 3.95e-25 kgIf Bfield = 0.750 T, find the separation distance after the isotopes complete half circular path.

9

Problem 5: calculate the time t for a particle of mass m to complete one circular orbit in a magnetic field.

HELICAL MOTION results when a initial velocity of charged particle is not perpendicular to Bfield

MAGNETIC FORCE EXERTED ON CURRENT CARRYING WIRE

For wire of length L, carrying current I in Bfield: F = qvBsin = ILBsin Current carrying wire experiences ZERO force when it parallel to Bfield.

Current carrying wire experience maximum force when perpendicular to Bfield.

10

Problem 6: A copper wire 0.150 m long with mass 0.0500 kg is suspended by 2 thin flexible wires. A Bfield [ 0.550T] is at right angles to the copper wire. Find the direction and magnitude of the electric current needed to levitate the copper wire.

LOOPS OF CURRENT AND MAGNETIC TORQUE:

In electric motors and generators,current carrying wires are in loops.

Rectangular current loop placed in uniform Bfield parallel to plane of loop

Loop Height=h loop width=w current=I

11

The horizontal portions of loop are // to Bfield so experience zero force

The two vertical components of loop experience opposite forces [current runs in opposite directions]

left side of loop: Bforce is into the page right side of loop: equal Bforce out of page

visualize the torque exerted by these forces around the center line of the loop.

= F r = ILB [r] + ILBr =2 IhB[w/2] = IB[hw]

= IBA Where A is the area of the loop.

If the rectangular wire loop is at angle wrt Bfield:

= IABsin units ofm

Torque exerted by general loop of area A that contains N turns: = NIABsin

Torque depends on number of turns, area of loop , strength of Bfield, and orientation wrt Bfield

Problem 7: A rectangular coil of 200 turns is 5.0 cm high and 4.0 cm wide. When coil is placed in 0.35 T Bfield the maximum torque is 0.22 N-m. what is the current in the coil?

The torque exerted by a Bfield has many applications such as electric motors and Galvanometers

12

Galvanometer: needle attached to coil of wire in Bfield will deflect when current runs through wire.Ampere’s law: Ampere’s law relates the magnetic field along a closed path to the electric current enclosed by that path.For any closed loop path, the sum of the length elements times the magnetic field in the direction of the length element is equal to the permeability times the electric current enclosed in the loop.Or: BIL=oIenclosed

o = permeability of free space o = 4X10-7 T-m/A

Bfield around long straight wire: concentric circles around the wire. Use right hand rule.

B = oI 2r o = permeability of free space o = 4X10-7 T-m/A

Problem 8: find the magnitude of Bfield 1 meter from a long straight wire carrying 1 A current.

13

Problem 9: a 52 microcoloumb particle moves parallel to a long wire at 720 m/sec. the separation between the particle and the wire is 13 cm. The force exerted on the particle is 1.4X10 -7 N.Find the magnitude of the Bfield at the location of the particle and the current in the wire.

Problem 10: two parallel wires separated a distance 22 cm carry currents in the same direction. The current in one wire is 1.5 A and the current in the other wire is 4.5 A. find the magnitude of the Bfield halfway between the wires.

Force between two parallel current-carrying wires: F = o I 1I2 L

2d

14

If currents in same direction, wires attract each other. If in opposite directions, wires repel each other

Current loops and solenoids: Bfield produced by circular current carrying wire loop resembles Bfield of a bar magnet.

For N circular loops of current-carrying wire B = N oI[2R]

N = number of turns

R = radius of loop

Magnetic forces between 2 current loops: loops behave like bar magnets, so if currents in same direction, loops will repel If currents in opposite direction loops will attract

A solenoid is a long wire wrapped in a helix. There is a strong uniform field inside the helix and a weak [almost zero] field outside the helix.

Bfield of a solenoid = o[N/L] I = o Ni Where n = N/L

Problem 11: a 200 turn solenoid is 20 cm long and carries a current of 3.25 A. Find the magnitude of the force exerted on a 15 microcoulomb charged particle moving at 1050 m/sec through the interior of the solenoid, at angle 11.5o wrt the solenoid’s core.

15

MAGNETIC FLUX AND FARADAY’S LAW

Oersted observed an electric current produces a magnetic field Michael Faraday showed a changing magnetic field can produces a current .

If a loop of wire is placed in a Bfield no current flows. But if the magnitude of the Bfield changes, then current flows.

Electric circuit: When a potential difference [from a battery, for example] causes current to flow through a wire loop.

With magnetic induction, emf is caused by the changing Bfield. This “voltage” is induced emf

Induced emf = potential difference created by a changing Bfield that causes a current to flow in a wire. Emf stands for electro-motive force but it is NOT a force.

For a loop of wire to “feel” the changing Bfield, some of the field lines must pass through it. The amount of Bfield that passes through a loop is called the magnetic flux.

Magnetic flux : the number of magnetic field lines that pass through an area.

= BcosA The units of magnetic flux are webers 1 weber= 1Wb = 1 T-m2

B = magnitude of Bfield A = area of region penetrated by Bfield.

If the Bfield is parallel to the Area, then magnetic flux = 0.

If the Bfield is perpendicular, then magnetic flux is maximum.

16

Magnetic flux depends on strength of Bfield, Area A and orientation of A and B.

Problem 12: consider a circular loop [radius = 2.50 cm] in a constant Bfield B= 0.625 T. Find the magnetic flux through this loop with the axis normal to the loop is oriented

a) 0o wrt the Bfield

b) 30o wrt the Bfield

c) 60o wrt the Bfield

d) 90o wrt the Bfield

Any change in the magnetic environment of a coil of wire will cause a voltage (emf) to be "induced" in the coil. No matter how the change is produced, the voltage will be generated. The change could be produced by changing the magnetic field strength, moving a magnet toward or away from the coil, moving the coil into or out of the magnetic field, rotating the coil relative to the magnet, etc.

Faraday’s law of induction: = -N /t induced emf in coil = number of loops in a coil

Magnitude of induced emf = N |/t | = N| final – initial| |tfinal –tinitial|

Problem 13: A bar magnet is moved rapidly toward a 40 turn circular coil of wire with radius 3.05 cm. The average value of Bcosover the area of the coil increases from 0.0125T to 0.450T in 0.250 seconds. The resistance of the wire is 3.55. Find the induced emf[use Faraday’s law] and the induced current [use Ohm’s law].

Applications of Faraday’s law: dynamic microphone electric guitar magnetic disk drive card reader

Lenz’s law: current always flows in a direction that opposes the change that caused it. 17

Or: An induced electromotive force generates a current that induces a counter magnetic field that opposes the magnetic field generating the currentOr: the direction of the induced current opposes the increase of flux.

Example: a)a wire loop near a region containing a Bfield as shown. Initially, the magnetic flux through the wire is zero.

a)

b)Now move the wire into the Bfield. As Bfield lines begin to pass through the loop. The magnetic flux is changing [increasing] and a current is induced in the wire. In which direction?

b)

c)After a while the entire loop enters the region. There is no more changing flux, because same number of lines pass through loop. So, no induced emf, current stops.

c)

d)When loop begins to pass out of the Bfield, magnetic flux begins to decrease and current again flows. Which way?

d)

Current now flows in opposite direction.

18

Problem 14: Consider a metal ring falling through a Bfield directed into the surface, toward a field-free region . According to Lenz’s law is the induced current cw or ccw?.

The induced current causes the falling ring to slow down.

Induced emf in a rectangular wire loop

Consider a rectangular loop being pulled through a uniform Bfield at constant speed v. The induced emf is found by = BLv where L is the length of the side of the rectangle that is perpendicular to the direction of motion. Problem 15: Consider a uniform B-field perpendicular into the surface as shown. All of these rectangles are 5cm X10cm, and are being pushed to the left at a constant speed of 2 cm/sec. At this moment, the loops are 25%, 50%, 75% or 100% within the B-field as shown. a) Rank these loops A,B,C,D,E,F,G,H,J and K from having greatest to least amount of induced emf. B) indicate direction CW or CCW that current flowing in any of these loops at this time.

Eddy Currents

19

Faraday’s Law implies that a changing magnetic flux produces an induced electric field even in empty space.

If a metal plate is inserted into this empty space the induced electric field produces electric currents in the metal. These induced currents are called eddy currents. 

If the induced currents are created by a changing magnetic field then the Eddy currents will be perpendicular to the magnetic field and flowing in circles if the B-field is uniform. 

These induced electric fields are very different from electrostatic electric fields of point charges. For one thing they do not began and end on charges but circle around on themselves in loops.

20