PHYSICS 3 (ELECTRICITY AND MAGNETISM) - CHAPTER 4 docx

If we place some compasses near a bar magnet, the needles will align themselves along the direction of the magnetic field, as shown in Fig.. 4.2 : Magnetic field of a bar magnet When two

Chapter 4 MAGNETISM

4.1 The Magnetic field

The most familiar source of magnetic fields is a bar magnet One end of the bar magnet is called the North pole and the other, the South pole If we place some compasses near a bar magnet, the needles will align themselves along the direction of the magnetic field, as shown in Fig 4.1 The observation can be explained as follows: A magnetic compass consists of a tiny bar magnet that can rotate freely about a pivot point passing through the center of the magnet When a compass is placed near a bar magnet which produces an external magnetic field, it experiences a torque which tends to align the north pole of the compass with the external magnetic field

Fig 4.1 Fig 4.2 : Magnetic field of a bar magnet When two magnets or magnetic objects are close to each other, there is a force that attracts the poles together When two magnetic objects have like poles facing each other, the magnetic force pushes them apart (Fig 4.3) Magnets also strongly attract ferromagnetic materials such as iron, nickel and cobalt

Fig 4.3 Like poles repel, opposite poles attract

Fig 4.4 Magnetic field lines : Magnetic field lines emanate primarily from the north pole of a magnet and curve around to the south pole

The Earth’s magnetic field behaves as if there were a bar magnet in it (Fig 4.5) Note that the south pole

of the magnet is located in the northern hemisphere

Fig 4.5 : Magnetic field of the Earth Fig 4.6:The iron filings suggest

the magnetic field line of a bar magnet 4.2 Motion of a Charged Particle in a Uniform Magnetic Field

1) The Lorentz force

Consider a test particle with charge q moving through the magnetic field B

r

with the velocity vr The Lorentz force is

B x v

FB

r v r

The SI unit for B is the Tesla (T) : 1T = 1 N/(Am) = 104 Gauss

The right-hand rule gives the direction of a vector resulting from the cross product of two other vectors

To find the direction of the resulting vector sweep the fingers of the right hand from the direction of the first vector to the direction of the second vector over the smallest possible angle between the vectors The direction in which the thumb points is the direction of the resulting vector

Fig 4.7 : The right-hand rule 2) A charged particle circulating in a magnetic field (Fig 4.8)

A charge particle with mass m and charge magnitude |q| moving with velocity vr perpendicular to a uniform magnetic field B

r

will travel in a circle of radius r

r

mv vB

| q

|

2

= ⇒ r =

B

| q

|

mv

The frequency of the revolution

f =

π

ω

2 = T

1

=

r 2

v

B

| q

|

4.3 Magnetic Force Acting on a Current-Carrying Conductor (Fig 4.9)

Consider a length L of the wire in Fig 4.9 The amount of charge moving through the wire

q = it = iL/v

(v : drift speed)

⇒ FB = qvBsin(ϕ) = iLBsin(ϕ)

(ϕ : angle v, B)

A straight wire carrying a current i in a uniform magnetic field experiences a sideways force

B x L i

FB

r r r

the direction of the length vector L

r

is that of the current i (The length vector L

r

has magnitude |L

r

| = L and is directed along the wire segment in the direction of the current.)

Exercise: The mass spectrometer is shown in the following figure in which it shows an arrangement used to measure the masses of ions An ion of mass m and charge +q is produced essentially at rest in source S, a chamber in which a gas discharged is taking place The ion is accelerated by potential difference V and allowed to enter a magnetic field B

r In the field it moves in a semicircle, striking a photographic plate at distance x from the entry slit Show that the ion mass m is given by

2 2

B q

8V

4.4 Torque on a Current Loop in a Uniform Magnetic Field

A coil (of area A and N turns, carrying current i) in a uniform magnetic field B

r

will experience a torque

τ

r

given by

B x

r r r

µ

=

here µr is the magnetic dipole moment of the coil, with magnitude µ = NiA and direction given by the right hand rule (grasp the coil with the fingers of the right hand in the direction of the current i, the thumb point to the direction of µr (Fig 4.10)

Fig 4.10

4.5 The Hall effect

Fig 4.11 : a strip of copper carrying a current i is immersed in a magnetic field B The charges (electrons) will experience a deflecting force FB Under the effect of the force FB, the electrons will be pushed toward the right edge of the strip, leaving uncompensated positive charges in fixed positions at the left edge An electric field E is produced within the strip, pointing from left to right The electric field exerts an electric force FE on each electron, tending to push it to the left An equilibrium is established when the electric force cancels the magnetic force

The Hall potential difference

V = Ed

When the electric force and the magnetic force are in balance

eE = evdB

Where vd is the drift speed : |vd| =

neA

i

A : cross-sectional area of the strip, n : number of charge per unit volume

Fig 4.11

4.6 The Biot-Savart Law

1) The magnetic field set up by a current-carrying conductor can be found from the Biot-Savart Law : The contribution d B

r

to the field produced by a current element idsr at a point P located a distance r from the current element is

3

o

r

r x s id 4 B d

r r r

π

µ

Here rr is a vector that points from the element to P The quantity µo = 4πx10-7 Tm/A ≈ 1.26x10-6 called the permeability constant

Fig 4.12 2) Magnetic Field of a Long Straight Wire (Fig 4.13)

Biot-Savart law

⇒

2

o

r 4

) sin(

ids dB

π

θ µ

with r = s +2 R2 and sin(θ) = sin(π-θ) =

2

2 R s

R +

r 4

) sin(

i 2

µ

0 2 2 3/2

) R s (

R 2

i

=

R 2

i

o

π

µ

3) Magnetic field due to a current in a circular arc of wire (Fig 4.14)

Arc-shaped wire with central angle Φ, radius R, center C, carrying current i

B =

R 4

i

o π

Φ µ

4) Force Between Two Parallel Currents

Two parallel wires carrying currents in the same direction attract each other (Fig 4.15)

•••• Ba is the magnetic field at wire b produced by the current in wire a

•••• Fba is the resulting force acting on wire b because it carries current in field Ba

In Fig 4.16, the system of two current carrying wires is viewed in the direction of the currents With the currents perpendicular to the plane of the drawing and directed "into" the plane, the magnetic field created

by current ia circulates along (is tangent in clockwise direction to) circles centered at current ia The figure shows the direction of this magnetic field Ba at the location of current ib

The magnitude of Ba at every point of wire b is :

Ba =

d 2

ia

o

π

µ

(4.9)

The force Fba on a length L of wire b due to the external magnetic field Ba is

a b

ba i L x B F

r r r

Since L and Ba are perpendicular to each other

Fba =

d 2

i

Lia b

o

π

µ

Parallel currents attract each other Antiparallel currents repel each other

4.7 Ampere’s Law

1) Ampere’s Law: Consider Fig 4.17

∫Bdsr

r

2) Magnetic field outside a long straight wire with current (Fig 4.18)

∫Bdsr

r

= 2πrB = µoi ⇒

r 2

i

π

µ

3) Magnetic field inside a long straight wire with current (Fig 4.19)

2

o

R 2

ir B

π

µ

4.8 The Magnetic Field of a Solenoid

1) Solenoid

Ampere’s law (Fig 4.20)

r

= Bh = µoienc = µonhi ⇒ B = µoni (4.16)

n : number of turns per unit length

2) Toroid

Ampere’s law (Fig 4.21)

∫Bdsr

r

= B2πr = µoienc = µoNi

r 2

Ni

o

π

µ

(4.17)

N : total number of turns

In contrast to the situation for a solenoid, B is not constant over the cross section of a toroid

3) Current-Carrying coil as a Magnetic Dipole (Fig 4.22)

Biot-Savart law ⇒⇒⇒

B =

( 2 2)3 / 2

2 o

z R 2

iR +

µ

(4.18)

4.9 Magnetic Flux Gauss’s Law in Magnetism

1) The magnetic flux ΦB through an area A in a magnetic field B

r

is defined as

ΦB = ∫B.dA

r r

where the integration is taken over the area

2) Gauss’s Law in Magnetism

The net magnetic flux through any (closed) Gaussian surface is zero

⇒ The simplest magnetic structure that can exist is a magnetic dipole Magnetic monopoles do not exist

4.10 Displacement Current and the General Form of Ampère’s Law

1) Maxwell’s law of induction : a changing electric flux induces a magnetic field B

r

∫Br.dsr = µoεo

dt

dΦE

(4.21)

Fig 4.23 : A circular parallel plate capacitor is being charged by a constant current Example : A parallel plate capacitor with circular plates of radius R is being charged as in Fig 12.1 Derive an expression for the magnetic field at radius r ≤ R Evaluate the field magnitude for r = R/5 = 11mm and dE/dt 1.5x1012 V/ms Derive an expression for the magnetic field at radius r > R

r ≤ R : 2πrB = µoεo

dt

dΦE

= µoεo(πr2)

dt

dE

⇒ B = µoεo

2

r dt dE

r > R : 2πrB = µoεo

dt

dΦE

= µoεo(πR2)

dt

dE

⇒ B = µoεo

r 2

R2 dt dE

2) Ampere-Maxwell law

Ampere’s law

Combining (4.21) and (4.22) yields Ampere-Maxwell law

∫Br.dsr = µoi +

dt

o o

Φ ε

3) Displacement current

The quantity

id =

dt

o

Φ

has the dimension of a current and is called the displacement current Rewrite (4.23)

Fig 4.24 : i = id

The displacement current id can be viewed as the continuation of the real current i (Fig.(12.2)) The magnitude and the direction of the magnetic field produced by the displacement current id is determined as the one of the real current i

Example : The circular parallel plate capacitor in previous example is being charged with a current i Determine the magnetic field B at a radius r from the center Assume that id is uniformly spread over the full plate area

∫Br.dsr = µoid

2

2

R

r

⇒ 2πrB =

2

2 d o

R

r i µ

2 d o

R 2

r i π

µ =

2

o

R 2

ir π µ (where the integration is taken over the circle of radius r.)

Problems

Magnetic field

4.1) A flexible wire, carrying a current i, passes between the pole faces of a magnet Under the influence of the magnetic field, the wire is deflected Determine the direction of the current i in each case (Fig P4.1)

Fig P4.1

4.2) In Fig P4.2, a metal wire of mass m = 25mg can slide with negligible friction on 2 horizontal parallel rails separated by distance d = 4cm The track lies in a vertical uniform magnetic field of magnitude 50mT At time t = 0, a source is connected to the rails, producing a constant current i = 10mA in the wire and rails (even as the wire moves) At t = 50ms, what are the speed and the direction of motion of the wire

4.3) An ion of mass m and charge q is produced in source S (Fig P4.3) The initially stationary ion is accelerated by the electric field due to a potential difference e The ion leaves S and enters a separator chamber in which a uniform magnetic field B is perpendicular to the path of the ion The magnetic field B causes the ion to move in a semicircle and thus strikes a detector at the bottom wall of the chamber Suppose that B = 80mT, e = 1000V, q = +1.6022x10-19C, x = 1.6254m What is the mass m of the individual ion ?

4.4) Magnetic levitation is used in high-speed trains Conventional electronmagnetic technology is used to suspend the train over the tracks; the elimination of rolling friction allows the train to achive very high speeds (in excess of 400km/h) The principle of magnetic levitation can be given as the following problem A straight horizontal copper rod carries a current of 50.0 A from west to east in a region between the poles of large electromagnet (Fig P4.4) In this region there is a horizontal magnetic field toward the north-east (that is, 45o north of east) with magnitude 1.20 T Find

a) The magnitude and direction of the force on a 1.00-m section of rod

b) If the horizontal rod is in mechanical equilibrium under the action of its weight and the magnetic force What is the mass of the horizontal rod?

c) While keeping the rod horizontal, how should it be oriented to maximize the magnitude of the force d) What is the force magnitude and the mass of the rod in case (c)

4.5) Two concentric, circular wire loops, of radii r1 = 12cm and r2 = 10cm, are located in an xy plane, each carries a clockwise current of 2A Find the magnitude of the net magnetic dipole moment of the system (Fig P4.5A) Repeat for the reversed current in the outer loop (Fig P4.5B)

A B Fig P4.6

Fig P4.5

4.6) Consider a rectangular coil of wire in a magnetic field as shown in Fig P4.6 The coil has height a and width b The current in the coil is i

a) Find the force on each side of the coil

b) As the rectangular wire rotates, the force on the sides AB and CD is non-zero Does this effect the rotation ?

c) Are the forces on sides BD and AC constant in magnitude throughout a given rotation ?

4.7) A solid metal cube of edge length d = 1.5cm, moving in the positive y direction at velocity v = 4m/s through a uniform magnetic field B = 0.05T in the positive z direction (Fig P4.5)

a) Which cube face is at a lower electric potential and which is at a higher electric potential ?

b) What is the potential difference between the faces of higher and lower electric potential ?

Magnetic field by an electric current

4.8) Find the magnetic field at point O in Fig P4.8 where OA = 15cm, OB = 20cm, θ = π/3 rad, I = 1A

4.9) Find the magnetic field at the center O of the semicircle in Fig P4.9 where L = 12cm, R = 10cm

4.10) A conducting rectangle MNPQ (Fig.P4.10), carrying current I2, is placed near a long wire carrying current I1 Find the net force on the rectangle due to I1

4.11) Find the magnetic field at point P in Fig P4.11

Displacement Current and the General Form of Ampère’s Law

4.12) The magnitude of the electric field between the two circular parallel plates is E = 4x105 – 6x104t V/s (Fig P4.12) The plate area is 4x10-2 m2 Determine

a) the magnitude and the direction of the displacement current between the plates

b) the magnitude and the direction of the induced magnetic field

Fig P4.12 Fig P4.13 4.13) Two wires, parallel to a z axis and a distance 2r apart, carry equal currents i in opposite directions as shown in Fig P4.13 A circular cylinder of radius r/2 and length L has it axis on the z axis, midway between the wires Use Gauss’ law for magnetism to derive an expression for the net outward magnetic flux through the half of the cylindrical surface above the x axis (Hint : find the flux through the portion of the xz plane that lies within the cylinder.)

4.14) A capacitor C with circular plates of radius b The distance between the two plates is d Initially the capacitor is charged to a voltage Vo At t = 0 the switch is closed and the capacitor discharges through the resistor R (Fig P4.14)

a) Find the charge Q as a function of time of the capacitor

b) Find the electric field E, the magnetic field B and the displacement current id between the capacitor plates

4.15) The capacitor C in Fig P4.15 has circular plates of radius b The space d between the two plates is small compared to b so that we can ignore the fringing effects Initially C is uncharged At t = 0 the switch K is closed and the capacitor charges through the resistor r

a) Find the potential difference V and the current i of the circuit

b) Find the electric field E, the magnetic field B and the displacement current id between the capacitor plates

4.16) Two square conducting loops carry currents of 5.0 A and 3.0 A as shown in Fig P4.16 What is the value of the line integral ∫Bdsr

r

for each of the two closed paths shown?

Fig P4.16