lecture 9-1 lorentz force - physics.purdue.edu

Lecture 9Lecture 9--11Lorentz Force

Let denote the electric and magnetic vector fields. The force actingon a point charge q, moving with velocity fields is:

and E B F in the superimosed v E B+

{ }F q E v B= + ×

This is called the Lorentz force equation.

Velocity Selector

Assume that the electric field is created by a parallel plate capacitor pointingalong the −Y axis and the magnetic field along the −Z axis as shown in the figure below. Since the moving charged particle is negative:

EB

(up)

(down)

Ey

By

F qE

F qvB

= +

= −

Lecture 9Lecture 9--22

When the electric and magnetic forces balance

0qE qvBEvB

− =

=

X

Y


Magnetic Mass spectrometer

When an ion of unknown mass enters the homogeneous magnetic field of the Magnetic spectrometer it executes a circular path. The measured radius of the circle depends on its mass m and its velocity . Thus v also has to be measured. (See Lec.8, page 12)

BB

v


Physics 219 – Question 1 – February 08, 2012.

a) Increase Eb) Increase Bc) Turn B offd) Turn E offe) Nothing

A proton (charge +e) comes horizontally into a region of perpendicularly crossed, uniform E and B fields as shown. In this region, it deflects upward as shown. What can you do to change the path so it remains horizontal through the region?

e+

Lecture 9Lecture 9--55Magnetic Force on a Current

• Consider a straight current-carrying wire in the presence of a magnetic field B.

• There will be a force on each of the charges moving in the wire. What will be the total force F on a length L of the wire?

• Current is made up of n charges/volume, each carrying charge q < 0 and moving with velocity vd through a wire of cross-section A.

qv B×• Force on each charge =

( )F n AL qv B= ×• Total force:

F iL B= ×

A

• On the next page we show that the productnAvq

is equal to the current flowing in the wire.


In time Δt, all the free charges in the shaded volume pass through A.If there are n charge carriers per unit volume, each with charge q, thetotal free charge in this volume is isthe drift velocity of the charge carriers.

, where d dQ qnAv t vΔ = Δ

Lecture 9Lecture 9--77Magnetic Force on a Current Loop

– Force on top path cancels force on bottom path (F = IBL)

– Force on right path cancels force on left path. (F = IBL)

loop

loop

I L B

I L B

F = ×

⎛ ⎞= ×⎜ ⎟

⎝ ⎠

∑

∑

Force on closed loop current in uniform B?

Uniform B exerts no net force on closed current loop.

closed loop

0=

Lecture 9Lecture 9--88 Magnetic Torque on a Current Loop

• If B field is ⊥ to plane of loop, the net torque on loop is 0.

r Fτ = ×definition of torque

abut a chosen point

so that n is twisted to align with B

n

n

magnetic moment direction

• If B field is // to plane of loop, the net torque on loop is maximum.

a

b

b

Lecture 9Lecture 9--99Calculation of Torque

2 sin2br F Fτ τ θ= × → = ⋅ ⋅∑

Thus: sinsin

Iab BIA B

τ θθ

= ⋅ ⋅=

• Note: if loop ⊥ B, θ = 0 and τ = 0. Maximum torque occurs when plane of the loop is parallel to B.

• Suppose the coil has width b (the side we see) and length a(into the screen). The torque about the center is given by:

μmagnetic moment

(Simple Approach)

The magnitude of magnetic force F acting on a wire of length a carrying current I in magnetic field B is: F IaB=

area of loop

Lecture 9Lecture 9--1010Calculation of Torque

(General Approach)

For reference : Giambattista, Vol. 2, Ch. 19., page 719., prob. 47.

It can be shown that the magnetic moment of planar loop of any shape of area A carrying a current I is: NIAμ =N denotes the number of turns.The magnetic torque in magnetic field is:

(19.13 )B NIA B bτ μ= × = ×The direction of is can be determined from the current direction in the loop and the right hand rule.

A

The direction of is along the rotation axis. The direction can beobtained by the right hand rule.

τ

Lecture 9Lecture 9--1111Sources of Magnetic Fields

70 2

N4 10A

πμ − ⎛ ⎞= × ⎜ ⎟⎝ ⎠

where the magnetic permeabilityconstant μ0 is

/T m A⋅also

0

2IBr

μπ

=B

r

magnetic field circulates around wire.

Since we could not find magnetic monopole, Gausstheorem for field gives:

0netB

S

B dS q⋅ = =∫However a French contemporary of Oersted,Ampere, noted that an infinite straight currentcarrying wire creates a circulating magnetic fieldaround the wire, where:

In 1820 Hans C. Oersted atUniversity of Copenhagenobserved that electric currentcreates magnetic field andlightning magnetizes iron.

0

2IBr

μπ

=

B


• Moving point charge

• (Bits of) current

I

How do we calculate B due to I?

by using calculus, or

Ampère’s Law (sometimes)

dB

In general: Biot – Savart Law

034

idl rdBr

μπ

×=

A field created by a element of a wire which conducts i current, at a distant point P from .

dldl


Circular Loop Current is a Magnetic Dipole

0( )2

IB at centerr

μ=

multiple loops ⇒

front

We have already discussed that a



A loop of wire lies in the x-y plane. This loop has a current I that circulates clockwise as viewed from above (from +z toward -z). What is the direction of the magnetic field at point A?

a) along +xb) along -yc) along +zd) along -ze) none of the above

z

x

y

A

I

Lecture 9Lecture 9--1616Solenoid’s B field synopsis

// to axis

• Long solenoid (R<<L):

B inside solenoid

B outside solenoid nearly zero

(not very close to the ends or wires)

Solenoid’s B field Bar magnet’s B field

L

R

Lecture 9Lecture 9--1717Ampere’s Law in Magnetostatics

The sum of the product of B|| (magnetic field projected along a path) and Δl (the path length) along a closed loop, Amperianloop, is proportional to the net current Inet encircled by the loop,

|| 0 netloo lp

l BB d Il μΔ = ⋅ =∫∑

• Choose a direction of summation.

• A current is positive if it flows along the RHR normal direction of the Amperian loop, as defined by the direction of summation.

0 1 2( )i iμ −

Δl

70 2

N4 10A

πμ − ⎛ ⎞= × ⎜ ⎟⎝ ⎠


// to axisB inside solenoid

B outside solenoid nearly zero

(not very close to the ends or wires)n windings per unit length

B

r

Long straight wire:

00

22 rB I IB

rπμ μπ = ∴ =

Long solenoid:

|| 0 netloop

B l IμΔ =∑

0 0( )Bh nhI B nIμ μ∴ ==

Calculation of the Solenoid Magnetic FieldWith the help of Ampere’s Law:

Lecture 9Lecture 9--1919Two Parallel Currents

12 2 2 1F I L B I LB= × =

0 11 2

IBR

μπ

=

0 1 22 2

I IF LR

μπ

∴ =

2 0 1 2 1

2F I I FL R L

μπ

= =

The ampere is defined to be the constant parallel currents 1 m apartthat will produce the force between them of 2 x 10-7 N per meter.

Definition of charge 1 C (1 A for 1 Sec)

L

1 2

lecture 9-1 lorentz force - physics.purdue.edu

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