We have seen that an individual charged particle moving through a magnetic field feels a force on it. What about a group of charges moving in a wire in a B-field? Presumably a current carrying wire will feel the sum of all the individual forces on the electrons in it.

Consider a wire of length l carrying a current i in a magnetic field B.

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Note that if i is the direction of electron flow, the force on the wire is down

We may think of the wire as having a total moving charge q, traveling at a velocity v. Using this notation:

F = qvB
F = q(l/t)B
F = (q/t)lB
F = ilB
F = liB Remember, current = charge/time

14. What force is exerted on a 5 cm wire carrying a current of 2.2 amp in a magnetic field of 0.2 web/m²?

F = liB F = (.05)(2.2)(0.2) F = .0220 nt

15. A wire of length l and mass m carries a current i. What strength field, B, will be necessary to float the wire up off the ground?

FB = Fg liB = mg B = mg/ib

16. A wire of length l is suspended by a spring of stiffness k in a magnetic field B. How far will the spring stretch when a current i is passed through the wire?

FB = FS liB = kx x = liB/k

17. A wire of length l and mass m rests on two conducting rails as shown. The wire has a resistance R, and the rails are connected to the battery E. A magnetic field B exists perpendicular to the rails. What coefficient of friction is necessary to keep the wire from sliding?

Ff = FB µmg = liB µmg = l(E/R)B µ = lEB/mgR

18. A charged particle q is fired between two charged plates containing a downward electric field, E. A B-field out at right angles to the E-field forces the particle oppositely so that it ends up traveling in a straight line. What is the velocity of the particle?

The downward force, qE, is balanced by the upward electro-magnetic force, qvB. qE = qvB v = E/B

19. An electron passes straight through crossed E and B fields of 3.0 * 10⁴ nt/C and 0.20 web/m². It then enters a 0.35 web/m² B-field. What is the radius of its orbit?

To go straight through the E and B fields, we know qE = qvB or v = E/B. But in orbit, mv2/r = qvB2. mv2/r = qvB2 r = mv/qB2 r = mE/qB1B2 r = [(9.1 * 10-31)(3.0 * 104)]/[(1.6 * 10-19)(.2)(.35)] r = 2.44 * 10-6 m

20. Consider a wire of length l moving at velocity v through a magnetic field B.
a) What force does a charge q in the wire feel?
b) How much work is required to push a charge q from one end of the wire to the other?
c) What is the work/charge, or potential, between the ends of the wire?

a) F = qvB

b) Work = F * l Work = qvBl

c) V = W/q V = qvBl/q V = lvB

The solution to problem 20 is terribly important. With rare exceptions, V = lvB is the way electricity is generated in this world. Wires, thousands of them, are moved rapidly through magnetic fields, generating the electricity that runs out society.

21. A 12 cm long wire is pushed through a 0.2 web/m² field at 3.0 m/sec. What potential is generated between the ends of the wire?

V = lvB V = (.12)(3.0)(0.2) V = .0720 V note: The wire must be perpendicular to B and to its direction of motion.

22. A wire of length l₁ is moved at velocity v through field B₁. Another wire of length l₂ is in field B₂. What force does it feel?

The moving wire generates a potential E = l1vB1 which produces a current: i = E/R i = l1vB1/R F = l2iB2 F = (l1l2vB1B2)/(R)

23. A wire, moving through field B, is attached to an electron gun as shown. What is the velocity of a charged particle (q, m) after it leaves the gun?

The gun has a potential v = lvB, but: qV = (1/2)mv2 v = [(2qV)/m]1/2 v = [(2qlvB)/m]1/2

24. How fast must a 30 cm long wire move through a 1.2 web/m² field to charge the 300 µF capacitor with 4.4 * 10^-3 coulombs?

The potential generated (lvB) must equal the potential of the capacitor (q/c). lvB = q/c v = q/lBc v = (4.4 * 10-3)/[(.3)(1.2)(300 * 10-6)] v = 40.7 m/sec

25. A charge q is placed on capacitor C which is then discharged through resistor R as shown. What initial, momentary force is felt by the wire?

F = liB F = l(E/R)B l = [(q/c)/(R)]B l = lqB/Rc

26. A rectangular loop of wire measuring a in width and b in length carries a current i in a magnetic field of strength B. What torque is exerted on it?

Use your negative hand to prove to yourself that the top wire is pushed out of the screen while the bottom is pushed in. Each force is F = liB and exerts a torque: T = F * r T = liBr T = biB(a/2). Since there are two wires: T = 2[biB(a/2)] T = abiB

The effect shown in problem 26 is the principle behind essentially all electric motors. A torque is generated in a current-carrying loop in a magnetic field. The trick to making this practical is to devise a sliding contact that will cause the direction of the current to change when the wire loop is rotated 180°.

Before proceeding, you must know abut the concept of flux. I think of flux as a total quantity of magnetic field, or the total of all field lines in a region of space. These definitions lack the formality needed to solve problems, so the proper definition is:

flux = area * magnetic field strength
¤ = A * B

It is assumed in this definition that the plane of area A is perpendicular to the magnetic field B. After all, if the surface intercepting the flux were parallel to the field, there would be no field lines passing through it, and hence no flux.

Here are some pictures of the same amount of magnetic flux:

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They are shown with increasing field strength, but decreasing areas. In this way the total flux is kept constant.

27. A rectangular wire, measuring 8 cm by 15 cm is held perpendicular to a 0.12 web/m² field. What is the flux through the loop?

¤ = A * B ¤ = (.08 m)(.15 m)(.12 web/m2) ¤ = 1.44 * 10-3 webers

Look at your units!! That is what a weber is...a unit of magnetic flux. The more concentrated the flux, the stronger the magnetic field. That is, there are more webers per square meter.

28. A 3.4 * 10^-8 kg particle with a 7.1 * 10^-7C charge orbits in a 1.1 web/m² field at 87 m/sec. What flux is contained within the path of its orbit?

First, the orbital radius: mv2/r = qvB r = mv/qB ¤ = AB ¤ = (3.14)(mv/qB)2B ¤ = (3.14)[ [3.4 * 10-8)(87)]/[(7.1 * 10-7)(1.1)] ]2(1.1) ¤ = 49.6 web

29. A magnetic field of 2.2 web/m² is parallel to the x-axis of a coordinate system as shown. A wedge-shaped frame of wire is placed in the field.
a) What is the flux through surface ABCD?
b) What is the flux through ADEF?
c) ....through ABF?
d) ....through BCEF?

a) ¤ = AB ¤ = (.12)(.10)(2.2) ¤ = .0264 web

b) ¤ = 0 note: ABEF is parallel to the field

c) ¤ = 0 note: No field lines penatrate the surface

d) ¤ = AB ¤ = (.10)(.24)cos(60)(2.2) ¤ = .0264 web note: cos(60) gets the projection of BCEF perpendicular to the field. It's important to note, that any line passing through BCEF will also pass through ABCD. Thus ¤d = ¤a.

30. A rectangular loop of wire measures 30 cm by 40 cm. It is placed in a magnetic field as shown. (You are seeing an edge-on view.) If the field is 0.2 web/m², what is the flux through the loop in positions (a), (b), and (c)?

a) ¤ = AB ¤ = (.3)(.4)(.2) ¤ = .024 web

b) ¤ = ABcosØ ¤ = (.024)cos(40) ¤ = .0184 web

c) ¤ = ABcosØ ¤ = 0

31. A circular loop of wire of radius r is placed in a field of strength B at angle Ø as shown. What is the flux through the loop?

the area of the loop = (3.14)r2, but its apparent area as seen by the field is reduced by a factor of cosØ. Thus: ¤ = ABcosØ ¤ = (3.14)r2BcosØ

Now let us return to the generation of potential. We saw in problem 20 that for a straight wire sweeping through a field B, the potential generated is V = lvB. Notice that l is in a vertical direction while v is horizontal. This means that the wire sweeps out a certain area and therefore cuts through some amount of flux each second.

V = lvB
V = l(x/t)B
V = lxB/t
V = ¤/t

We have, then, a new formulation of the way potential is generated in a wire. It is called Faraday's Law of induction.

E = Ǥ/Çt or the rate at which flux is changing.

As we shall see, this is a more powerful and wide-ranging way to state the law. It can be remembered most easily in its calculus notation, E = Ǥ/Çt, pronounced ";eee equals deee feee deee teee";.

32. A wire sweeps through 1.2 webers in 3.0 seconds. What potential is generated along its length?

E = Ǥ/Çt E = (1.2 web)/(3.0 sec) E = 0.40 V

33. A circular loop of wire with 9.0 cm radius is in a field of 0.85 web/m² strength. The field is reduced to zero in 0.12 seconds, inducing what potential in the loop?

E = Ǥ/Çt E = ÇB/t E = [(3.14)(.09)2(.85)]/(.12) E = 0.180 V

34. A rectangular loop of wire measuring 20 cm by 30 cm is placed in a magnetic field 0.09 web/m². The loop is rotated 90° in 0.06 seconds. What potential is generated?

E = Ǥ/Çt E = AB/t E = [(.2 * .3)(.09)]/(.06) E = .090 V

35. A circular loop of wire of radius r contains N turns of wire. It is placed in a field B and then rotated 90° in time t. If the total resistance of the wire is R, what current is generated in it?

¤ = (3.14)r2B E = Ǥ/Çt E = (3.14)r2B/t E = N[Ǥ/Çt] E = N(3.14)r2B/t i = E/R i = [N(3.14)r2B]/(Rt) note: (3.14)r2B/t is the potential generated in every turn of the wire.

36. A circular loop of wire has 35 turns in it. It is placed in a field of strength 0.04 web/m² and rotated 90° in 0.03 seconds. The radius of the loop is 12 cm and it is connected to a 200 µF capacitor. What charge is generated on the capacitor?

E = N(Ǥ/Çt) E = 35[ [(3.14)(.12)2(.04)]/(0.3) ] E = 2.11 V q = CV q = (200 µ)(2.11) q = 422 µC

37. A rectangular loop of wire has dimensions a and b and has N turns. A wire of length l and mass m rests on frictionless rails as shown. If the loop is rotated 180° in time t, what will be the acceleration of the wire?

a = F/m a = liB2/m a = [l(E/R)B2]/m a = (lEB2)/(Rm) E = N(Ǥ/Çt) E = N[(abB1)/(t/2)] a = (lEB2)/(Rm) a = (2lNabB1B2)/(Rmt) note: t/2 is the time to go from max to zero, a rotation of 90°.

38. A wire of mass m is loosely connected to two vertical rails and is allowed to fall through magnetic field B.
a) What is the direction of current through the wire generated by the wire's fall?
b) This current generates a force on the wire itself. What is the direction of this force?
c) When the wire drops at velocity v, what force is generated by the current in the wire?
d) What is the terminal velocity of the wire?

a) Electron flow in the wire is to the left.

b) A leftward flow generates an upward force.

c) F = liB F = l(E/R)B F = l(lvB/R)B F = l2vB2/R

d) Terminal velocity is reached when the upward force balances the weight of the wire. mg = l2vB2/R v = (mgR)/(l2B2)