B here τ== = = Fd iBa b iBab iBA () - University of Florida · ÎBut there is a net torque! ......

PHY2054: Chapter 19 21

a

a

bb

Torque on Current LoopConsider rectangular current loop

Forces in left, right branches = 0 Forces in top/bottom branches cancelNo net force! (true for any shape)

But there is a net torque!Bottom side up, top side down (RHR)Rotates around horizontal axis

μ = NiA ⇒ “magnetic moment” (N turns)True for any shape!!Direction of μ given by RHR

B

( )Fd iBa b iBab iBAτ = = = =Plane normal is ⊥ B here


General Treatment of Magnetic Moment, Torque

μ = NiA is magnetic moment (with N turns)Direction of μ given by RHR

Torque depends on angle θ between μ and B

sinBτ μ θ=


Torque ExampleA 3-turn circular loop of radius 3 cm carries 5A current in a B field of 2.5 T. Loop is tilted 30° to B field.

Rotation always in direction to align μ with B field

30°

( )22 23 3 5 3.14 0.03 0.0339A mi rμ π= = × × × = ⋅

sin30 0.0339 2.5 0.5 0.042 N mBτ μ= = × × = ⋅


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

Force is always ⊥ velocity and ⊥ BPath will be a circle. F is the centripetal force needed to keep the charge in its circular orbit. Let’s calculate radius R

FFv

R

v

B

qF

v


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

BqF

v

2mv qvBR

=mvRqB

=


Cosmic Ray ExampleProtons with energy 1 MeV move ⊥ earth B field of 0.5 Gauss or B = 5 × 10-5 T. Find radius & frequency of orbit.

212

2KK mv vm

= ⇒ =

2mv mKReB eB

= =

( )( )6 19 13

27

10 1.6 10 =1.6 10 J

1.67 10 kg

K

m

− −

−

= × ×

= ×

( )1

2 2 / 2v v eBf

T R mv eB mπ π π= = = = 760Hzf =

2900mR =

Frequency is independent of v!


Helical Motion in B FieldVelocity of particle has 2 components

(parallel to B and perp. to B)Only v⊥ = v sinφ contributes to circular motionv|| = v cosφ is unchanged

So the particle moves in a helical pathv|| is the constant velocity along the B fieldv⊥ is the velocity around the circle

v v v⊥= +

mvRqB

⊥=

Bv

φ

v⊥

v||


Helical Motion in Earth’s B Field


Magnetic Field and WorkMagnetic force is always perpendicular to velocity

Therefore B field does no work!Why? Because

ConsequencesKinetic energy does not changeSpeed does not changeOnly direction changesParticle moves in a circle (if )

( ) 0K F x F v tΔ = ⋅Δ = ⋅ Δ =

v B⊥


Magnetic Force Two particles of the same charge enter a magnetic field with the same speed. Which one has the bigger mass?

ABBoth masses are equalCannot 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 xA B

mvRqB

=

Bigger mass meansbigger radius


Mass Spectrometer


Mass Spectrometer OperationPositive ions first enter a “velocity selector” where E ⊥ B and values are adjusted to allow only undeflected particles to enter mass spectrometer.

Balance forces in selector ⇒ “select” v

Spectrometer: Determine massfrom v and measured radius r

/qE qvBv E B

==

1

2

11

22

m vrqBm vrqB

=

=


Mass Spectrometer Example A beam of deuterons travels right at v = 5 x 105 m/s

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

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

5 5/ 10 /5 10 0.2T

eE evB

B E v

=

= = × =

( )( )5

62

1 5 10 1.5 10 Hz2 6.28 5.2 10

vfT Rπ −

×= = = = ×

×

( )( )( )( )

27 522

19

3.34 10 5 105.2 10 m

1.6 10 0.2mv mvevB RR eB

−−

−

× ×= ⇒ = = = ×

×

B here τ== = = Fd iBa b iBab iBA () - University of Florida · ÎBut there is a net torque! ......

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Transcript of B here τ== = = Fd iBa b iBab iBA () - University of Florida · ÎBut there is a net torque! ......