Physics 1202: Lecture 11Today’s Agenda

• Announcements:– Lectures posted on:

www.phys.uconn.edu/~rcote/

– HW assignments, solutions etc.

• Homework #4:Homework #4:– Not this week ! (time to prepare midterm)Not this week ! (time to prepare midterm)

• Midterm 1:– Friday Oct. 2

– Chaps. 15, 16 & 17.

Lorentz Force• The force F on a charge q moving with velocity v

through a region of space with electric field E and magnetic field B is given by:

F

x x x x x x

x x x x x x

x x x x x xv

B

q

v

B

qF = 0

v

B

qF

Units: 1 T (tesla) = 1 N / Am

1G (gauss) = 10-4 T

Lawrence's Insight"R cancels R"

• We just derived the radius of curvature of the trajectory of a charged particle in a constant magnetic field.

• E.O. Lawrence realized in 1929 an important feature of this equation which became the basis for his invention of the cyclotron.

• R does indeed cancel R in above eqn. So What??– The angular velocity is independent of R!!

– Therefore the time for one revolution is independent of the particle's energy!

– We can write for the period, T=2/ or T = 2m/qB– This is the basis for building a cyclotron.

• Rewrite in terms of angular velocity !

The Hall Effect

cd

l

a

c

B

BI

I

-

vd F

Hall voltage generatedacross the conductor

qEH

Force balance

Using the relation between drift velocity and current we can write:

Magnetic Force on a Current• Consider a 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?

• Suppose current is made up of n charges/volume each carrying charge q and moving with velocity v through a wire of cross-section A.

N S

Simpler: For a straight length of wire L carrying a current I, the force on it is:

• Force on each charge =

• Total force =

• Current =

or

Lecture 11, ACT1• A current I flows in a wire which is formed in the

shape of an isosceles triangle as shown. A constant magnetic field exists in the -z direction.

– What is Fy, net force on the wire in the y-direction?

(a) Fy < 0 (b) Fy = 0 (c) Fy > 0x

y

Magnetic Force on a Current Loop

x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x

B

i

• Consider loop in magnetic field as on right: If field is to plane of loop, the net force on loop is 0!

• If plane of loop is not to field, there will be a non-zero torque on the loop!

B

x

.FF

F

F

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

FF

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

Calculation of Torque• Suppose a square wire loop has width

w (the side we see) and length L (into the screen). The torque is given by:

rxF

• Note: if loop B, sin = 0 = 0

maximum occurs when loop parallel to B

since: A = wL = area of loopr

F

B

x

.FF w

Magnetic Dipole Moment• We can define the magnetic dipole moment of a current

loop as follows:

direction: to plane of the loop in the direction the thumb of right hand points if fingers curl in direction of current.

• Torque on loop can then be rewritten as:

• Note: if loop consists of N turns, = N A I

magnitude: A I

A I Bsin

B

x

.FF w

Electric Dipole Analogy

E

.

+q

-q

p

F

F

B

x

.FF

w

(per turn)

Lecture 11, ACT 2A rectangular loop is placed in a uniformmagnetic field with the plane of the loopparallel to the direction of the field. If acurrent is made to flow through the loop

inthe sense shown by the arrows, the fieldexerts on the loop:

A) a net force. B) a net torque. C) a net force and a net torque.

D) neither a net force nor a net torque.Solution

1. Bottom part force is out of page. Sides force = 0. Top force into page.

2. Sum of all forces is zero.

3. But torques, = r x F, not zero, I.e. loop will rotate.

4. Answer is B.

Lecture 11, ACT 3• A circular loop of radius R carries current I as

shown in the diagram. A constant magnetic field B exists in the +x direction. Initially the loop is in the x-y plane.

– The coil will rotate to which of the following positions?

(a) (b) (c) It will not rotate

Calculation of Magnetic Field

• Two ways to calculate the Magnetic Field:

• Biot-Savart Law:

• Ampere's Law

• These are the analogous equations for the Magnetic Field!

"Brute force" I

"High symmetry"

0= 4X 10-7 T m /A: permeability (vacuum)

Magnetic Field of Straight Wire

Direction of B:right-hand rule

Lecture 11, ACT 3• I have two wires, labeled 1 and 2, carrying equal

current, into the page. We know that wire 1 produces a magnetic field, and that wire 2 has moving charges. What is the force on wire 2 from wire 1 ?

(a) Force to the right (b) Force to the left (c) Force = 0

Wire 1

IX

Wire 2

IX

By right hand rule, B at wire 2 due to wire 1 is down.

B

The current direction is into the page.

From last lectures, F = I L x B, cross product to left

F

Force between two conductors

• Force on wire 2 due to B at wire 1:

• Total force between wires 1 and 2:

• Force on wire 2 due to B at wire 1:

• Direction:attractive for I1, I2 same direction

repulsive for I1, I2 opposite direction

Circular Loop

x

•

z

R

R

• Circular loop of radius R carries current i. Calculate B along the axis of the loop:

r B

r

z

B

• Symmetry B in z-direction.

>

>I

• At the center (z=0): z>>R:

• Note the form the field takes for z>>R: for N coils

Lecture 11, ACT 4• Equal currents I flow in identical

circular loops as shown in the diagram. The loop on the right (left) carries current in the ccw (cw) direction as seen looking along the +z direction.– What is the magnetic field Bz(A)

at point A, the midpoint between the two loops?

(a) Bz(A) < 0 (b) Bz(A) = 0 (c) Bz(A) > 0

• The right current loop gives rise to Bz <0 at point A.• The left current loop gives rise to Bz >0 at point A.• From symmetry, the magnitudes of the fields must be equal.• Therefore, B(A) = 0!

Lecture 17, ACT 3• Equal currents I flow in identical

circular loops as shown in the diagram. The loop on the right (left) carries current in the ccw (cw) direction as seen looking along the +z direction.

(a) Bz(B) < 0 (b) Bz(B) = 0 (c) Bz(B) > 0

– What is the magnetic field Bz(B) at point B, just to the right of the right loop?

• The signs of the fields from each loop are the same at B as they are at A!

• However, point B is closer to the right loop, so its field wins!

B Field of a Solenoid

• A constant magnetic field can (in principle) be produced by an sheet of current. In practice, however, a constant magnetic field is often produced by a solenoid.

• If a << L, the B field is to first order contained within the solenoid, in the axial direction, and of constant magnitude. In this limit, we can calculate the field using Ampere's Law.

L• A solenoid is defined by a current I flowing

through a wire which is wrapped n turns per unit length on a cylinder of radius a and length L.

a

B Field of a Solenoid

• To calculate the B field of the solenoid using Ampere's Law, we need to justify the claim that the B field is 0 outside the solenoid.

• To do this, view the solenoid from the

side as 2 current sheets.

xxx xx

•• • ••• The fields are in the same direction in the

region between the sheets (inside the solenoid) and cancel outside the sheets (outside the solenoid).

(n: number ofturns per unitlength)

Toroid• Toroid defined by N total turns

with current i.

• B=0 outside toroid!

• B inside the toroid.

x

x

x

x

x

x

x

x

x x

x

x x

x

x

x

•

•

•

•

• •

•

••

•

• •

•

•

••

r

B

Magnetism in Matter• When a substance is placed in an external magnetic field Bo,

the total magnetic field B is a combination of Bo and field due to magnetic moments (Magnetization; M):

– B = Bo + oM = o (H +M) = o (H + H) = o (1+) H

» where H is magnetic field strength is magnetic susceptibility

• Alternatively, total magnetic field B can be expressed as:– B = m H

» where m is magnetic permeability» m = o (1 + )

• All the matter can be classified in terms of their response to applied magnetic field:

– Paramagnets m > o

– Diamagnets m < o

– Ferromagnets m >>> o