Week 8 - Magnetic Field and Magnetic Forces

Ampere was the Newton of Electricity. James C. Maxwell

Exercise 8.1: Discussion Questions a) Can a charged particle move trough a magnetic field without experiencing any force? If so, how? If not, why not? Solution: Yes it can. Looking at the force law Fb = qv × B,

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the magnetic force is zero if v is parallel to B. I.e. the magnetic force is zero if a charged particle travel along the field lines. Particles with nonzero components of velocity at nonparallel with B end up spiralling around the field lines. b) If the magnetic field do no work, how can it have a significant effect on a particles motion? Are there any other examples for which a force has a significant effect on the motion without doing any work? Solution: The answer is that it can, without doing any work, change the direction of the velocity and therefore it’s motion. If the particle only has changes to it’s velocity and not it’s speed the kinetic energy does not change and thus no work is done. Another example of this is the familiar normal force from mechanics. The normal force on a roller coaster cart exerted by the rails does no work, but it certainly does affect the roller coaster carts motion! c) A student claims that if lightning strikes a metal flag pole, the force exerted by the earths magnetic field on the current flowing trough the pole can be enough to bend it. Typical lightning currents are of the order of 104 to 105 amperes. Is the student’s opinion justified? Solution: This calls for a rough estimate. The magnetic force magnitude on a straight rod is given by F = ILB, so assuming the worst possible scenario of a 105 amperes current and a 10 m rod we get a force on the order of

F ∼ 105 A × 10 m × 10−5 T ∼ 10 N. 1

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A force of the order of 10 N is not negigble. Remember this was just a rough estimate, so this calls for a more detailed calculation taking the material of the rod into account etc. The student could be right. d) A loudspeaker should never be placed in next to a computer monitor or a TV screen. Why not? Solution: Speakers contain powerful magnets which can permanently mess up the picture on a CRT dispaly. On an LCD or plasma display, the effect is less. However, even for the latter two types, there is chance that something would be magentized.

Exercise 8.2: Gauss’ Law for Magnetism In a certain region of space, the magnetic field B is not uniform. The magnetic field has both a zcomponent and a component that points radially away from the z-axis. The z-component is given by Bz (z) = βz, where β > 0. The radial component Br depends only on r, the radial distance from the z-axis. a) Use Gauss’ law for magnetism to find the radial component Br as a function of r. Hint: Try a clindrical Gaussian surface of radius r concentric with the z-axis with one end at z = z0 and the other at z. Solution: We use the hint and align a gaussian cylinder coaxial with the z-axis. This is shown in figure 1. Now as also shown in this figure the B-field can be decomposed in it’s z-component and it’s radial component and only the radial component Br will contribute to the flux trough the surface of the cylinder (not including the top and bottom surface where only the z-component will contribute to the flux). Further more since the radial field only depend on r it will be uniform along this surface and this is also the case for the bottom and top surface of the cylinder. We compute the flux

I

B · dA = 2πr(z − z0 )Br + βzπr2 − βz0 πr2 = 2πr(z − z0 )Br + βπr2 (z − z0 )

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where the last two terms corresponds to the flux from the top and bottom suface. Now by Gauss’ law for magnetism, this has to be zero, so we can solve for Br which yields

1 Br = − βr. 2

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1 Br = − βr. 2

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Answer:

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Figure 1

Exercise 8.3: Mass Spectrograph A mass spectrograph is used to measure the masses of ions, or to separate ions of different masses. In one design for such an instrument, ions with mass m and charge q are accelerated trough a potential difference V . They then enter a uniform magnetic field that is perpendicular to their velocity, and they are deflected in a semicicular path of radius R. A detector measures where the ions complete the semicircle and from this it is easy to cacluate R. The situation where a source S is emitting ions is shown in figure 2. In the red region there is a potential difference, while in the blue region there is a uniform magnetic field. a) Derive the equation for calculating the mass of the ion from measurements of B, V , R and q. Solution: The speed of the ions when they enter the region of a uniform magnetic field is found by energy conservation;

1 qV = mv 2 ⇒ v = 2

Week 8 – October 24, 2011

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r

2qV . m

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compiled October 22, 2012

Figure 2

When the ion enter this region it will feel a magnetic force

Fb = qv × B.

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Now this force is always perpendicular to v, so the acceleration is always perpendicular to v. This is a characteristic of circular motion, so that the ion will start going in a circle at once it enters the region of uniform B. Furthermore B is always perpendicular to v so the magnitude is

Fb = qvB = ma = m

v2 . R

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Thus

r qBR = mv = m

p 2qV = 2qV m m

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or equivalently

m=

qB 2 R2 . 2V

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m=

qB 2 R2 . 2V

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Answer:

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b) What potential difference V is needed so that singly ionized 0.150 T magnetic field?

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C atoms will have R = 50.0 cm in a

Solution: Rearranging the formula found in (b) we find

V =

qB 2 R2 2m

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The mass of Carbon-12 is 12 u and if it’s singly ionized it obtains a charge e. Thus the necessary potetial is


2 1.6 × 10−19 × (0.150) × 50.0 × 10−2 V = 22.5904 kV. V = 2 × 12 × 1.66 × 10−27

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2 1.6 × 10−19 × (0.150) × 50.0 × 10−2 V = V = 22.5904 kV. 2 × 12 × 1.66 × 10−27

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Answer:

c) Suppose the beam consists of a mixture of 12 C and 14 C ions. If V and B have the same values as in part (b), calculate the separation of these two isotopes at the detector. Do you think that this beam separation is sufficient for the two ions to be distinguished? Solution: Carbon-14 has a mass of 14 u. Again rearranging the formula in (a) and (b) differences in radii of the two semicircles will be

s ∆R =

2m14 C V − qB 2

s

2m12 C V = 4 cm. qB 2

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This beam separation should allow for easy separation of the two ions.

Answer: ∆R = 4 cm

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.

Exercise 8.4: An Electromagnetic Rail Gun A conducting bar with mass m and length L slides over horizontal rails that are connected to a voltage source. The voltage source maintains a constant current I in the rails and bar, and a constant, uniform, Week 8 – October 24, 2011

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Figure 3

vertical magnetic field B fills the region between the rails. This is shown in figure 3 where the black arrows indicate the direction of the current trough the grey rails and the purple conducting bar, while the blue arrows (pointing out of the page) indicate the direction of the B-field. a) Find the magnitude and direction of the net force on the conducting bar. Ignore friction, air resistance, and electrical resitance. Solution: Let’s align the axis of the railgun along the x-axis. Then the force on the conducting bar is then given by

FB = IL × B = ILBˆi

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i.e. it is directed along the axis of the rail gun. Since we are ignoring all other forces this is the only force and thus the net force on the bar.

Answer: FB = ILBˆi.

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b) If the bar has mass m, find the distance d that the bar must move along the rails from rest to attain a speed v. Solution: The force found in (a) will provide a constant acceleration given by a = ILB/m. Now for constant acceleration the familiar relation between distance d and speed v is given by

2ad = v 2 − v02 .

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So starting from rest we get that distance needed to acquire a speed v is

d=

2mv 2 v2 = . 2a ILB

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Thus it would be good to go for a huge current and magnetic field when designing a rail gun. The length L can not be made that large for pracitcal purposes in addition to all the air resistance it would cause and the greater mass. Week 8 – October 24, 2011

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Answer: d=

2mv 2 . ILB

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c) It has been suggested that rail guns based on this prinicple could accelerate payloads into earth orbit or beyond. Find hte distance the bar must travel along the rail if it is to reach the escape speed for the eart (11.2 km/s). Let B = 1.0 T, I = 3.0 kA, m = 1.0 kg and L = 1.0 m. For simplicity assume the net force on the object is equal to the magnetic force, as in parts (a) and (b), even though gravity plays an important role in an actual launch into space. Solution: This would require a distance d of


2 1.0 × 11.2 × 103 m = 21281.6 m d= 2 × 3.0 × 103 × 1.0 × 1.0

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In other words a pretty long rail gun. But we have achieved more favourable numbers. The highest ever magnetic field produced in a lab is about 100 T and and we have certainly produced electrical currents of about 105 A. With these numbers our distance would be

d = 6.4 m

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which suggests creating such a rail gun MIGHT be possible?

Exercise 8.5: Magnetic Moment of the Hydrogen Atom In the Bohr model of the hydrogen atom, in the lowest energy state the electron orbits the proton at a speed of 2.2 × 106 m/s in a circular orbit of radius a0 = 5.3 × 10−11 m called the Bohr radius. a) What is the orbital period of the electron? Solution: Let r = a0 be the radius of the electron and let v be it’s velocity. The period must be the time for one revolution. Now the distance traveled for a revolution is the circumference 2πr, so the velocity of the electron can be expressed as

v=

2πr T

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T =

2πr . v

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such that the period is given by

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Answer: T =

2πr . v

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b) If the orbiting electron is considered to be a current loop, what is the current I? Solution: The current is the charge of the electron divided by the period.

I=−

ev 2πr

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I=−

ev 2πr

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Answer:

c) What is the magnetic moment of the atom due to the motion of the electron? Show that it is proportional to the orbital angular momentum of the electron. Solution: The magnetic moment is given by µ = IA where A is the area of the ’current loop’ represented by the electron orbit. Therfore  ev 
evr πr2 = − µ= − . 2πr 2

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Now for an electron moving in a circuiar radius the angular momentum is given by L = rme v such that

µ=−

eL . 2me

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Now this result turns out to be out by a factor 2 from the actual value which is obtained by relativistic quantum mechanics. In fact the experimental and theoretical agreement on this value taking quantum mechanics in to account is a staggering 14 desimal places!

Answer: µ=−

Week 8 – October 24, 2011

eL . 2me

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Exercise 8.6: Force on a Semicircular Wire Figure 4 shows a conducting wire which enters a region of a uniform magnetic field B pointing out of the page. In this region the wire takes the form of a semicirle. The direction of the current running trough it is shown by the arrows. Find the net force on the conducting wire.

Figure 4

Solution:

Figure 5

The wire only feels a force in the region where there is a magnetic field. An infinitesimal segment of wire feels the force dF = Idl × B. Now as shown in figure 5 the contribution to the y-component of this force is given by

dFy = |dF| sin θ

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where |dF| = IBdl = IBRdθ so that the magnitude of the entire y-component of force is given by

Z Fy =

π

Z dFy = IBR

sin θdθ = 2IBR.

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0

The x-component is given by

Z

π

Fx = IBR

cos θdθ = 0

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0

which we also could have seen from the symmetry the problem. Furhermore by the right hand rule this force must point in the negative y-direction. Therefore we conclude that

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F = −2IBRˆj.

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F = −2IBRˆj.

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Answer:

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