Magnetic Moment and Torque on Current Loops in a Magnetic Field - Prof. Paul Ralph Avery, Study notes of Physics

The concept of magnetic moment and torque on current loops in a magnetic field. It covers the principles of net force being zero but net torque existing, the direction of magnetic moment given by the right-hand rule, and the dependence of torque on the angle between magnetic moment and magnetic field. The document also includes examples of calculating torque for a circular loop and the trajectory of a charge in a constant magnetic field.

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Pre 2010

Uploaded on 09/17/2009

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PHY2054: Chapter 19 21
a
a
b
b
Torque on Current Loop
ÎConsider rectangular current loop
Forces in left, right branches = 0
Forces in top/bottom branches cancel
No net force! (true for any shape)
ÎBut there is a net torque!
Bottom side up, top side down (RHR)
Rotates around horizontal axis
Îμ= NiA “magnetic moment” (N turns)
True for any shape!!
Direction of μgiven by RHR
B
()
Fd iBa b iBab iBA
τ
== = =
Plane normal is B here
pf3
pf4
pf5
pf8
pf9
pfa
pfd

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PHY2054: Chapter 19

21

a

a

b

b

Torque on Current Loop

Î

Consider rectangular current loop

‹

Forces in left, right branches = 0

‹

Forces in top/bottom branches cancel

‹

No net force! (true for any shape)

Î

But there is a net torque!

‹

Bottom side up, top side down (RHR)

‹

Rotates around horizontal axis

Î

μ

= NiA

“magnetic moment” (N turns)

‹

True for any shape!!

‹

Direction of

μ

given by RHR

B

Fd

iBa b

iBab

iBA

Plane normal is

B here

PHY2054: Chapter 19

General Treatment of Magnetic Moment, Torque Î

μ

= NiA is magnetic moment (with N turns)

‹

Direction of

μ

given by RHR

Î

Torque depends on angle

θ

between

μ

and B

sin

B

τ

μ

θ

PHY2054: Chapter 19

x x x x x x x x x x x x x xx x x x x x x x x x x x x xx x x x x x x x x x x x x xx x x x x x x x x x x x x x

Trajectory in a Constant Magnetic Field

Î

A charge q enters B field with velocity v perpendicular to B. What path will q follow?

‹

Force is always

velocity and

B

‹

Path will be a circle.

F is the centripetal force needed to keep the

charge in its circular orbit. Let’s calculate radius R

F

F

v

R

v

B

q

F

v

PHY2054: Chapter 19

x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x

Circular Motion of Positive Particle

B

q

F

v

2

mv

qvB

R

=

mv

R

qB

=

PHY2054: Chapter 19

27

Helical Motion in B Field

Î

Velocity of particle has 2 components

‹

(parallel to B and perp. to B)

‹

Only v

= v sin

φ

contributes to circular motion

‹

v

||

= v cos

φ

is unchanged

Î

So the particle moves in a helical path

‹

v

||

is the constant velocity along the B field

‹

v

is the velocity around the circle

v

v

v

&

G
G
G

mv

R

qB

B

v

φ

v

v

||

PHY2054: Chapter 19

Helical Motion in Earth’s B Field

PHY2054: Chapter 19

Magnetic Force

Î

Two particles of the same charge enter a magnetic field with the same speed.

Which one has the bigger mass?

‹

A

‹

B

‹

Both masses are equal

‹

Cannot tell without more info

x x x x x x x x x x x xx x x x x x x x x x x xx x x x x x x x x x x xx x x x x x x x x x x xx x x x x x x x x x x xx x x x x x x x x x x x

A
B

mv

R

qB

Bigger mass meansbigger radius

PHY2054: Chapter 19

Mass Spectrometer

PHY2054: Chapter 19

Mass Spectrometer Example

Î

A beam of deuterons travels right at v = 5 x 10

5

m/s

‹

What value of B would make deuterons go undeflected through aregion where E = 100,000 V/m pointing up vertically?

‹

If the electric field is suddenly turned off, what is the radius andfrequency of the circular orbit of the deuterons?

5

5

0.2T

eE

evB

B
E

v

×

(

)(

)

5

6

2

10 Hz

v

f

T
R

×
×
×

(

)(

)

(

)(

)

27

5

2

2

19

m

mv

mv

evB

R
R

eB

×
×
×
×