In the last chapter we noted the relation F = qvB. By rearranging this to F/q = vB and noting that F/q is just the electric field, E, we get E = vB. In physical terms we are saying that an electric field is created by the motion or change of a magnetic field. From this it is not surprising to learn that moving electric fields or charges will produce a magnetic field. This symmetry causes physicists to speak of the electromagnetic force and not consider it as two separate forces.

The detection of a magnetic field by an electric current is easily accomplished by arranging small compasses around a long wire carrying a fairly hefty current of several amps:

The field, then, forms concentric circles of decreasing magnetic field strength as one moves from the wire. Quantitatively,

B = k(i/r) where k = 2.0 * 10^-7 web/amp-m

The direction of the magnetic field may be found by the famous Left-Hand Rule. (Left hand because we are dealing with negative electron flow.)

1. Find the direction of the field around this current-carrying wire.

IMAGE

2. Find the direction of the current flow that would produce the B-field shown.

B is into the screen above the wire, out of the screen below the wire. The wire must be grasped with the thumb (and electron flow) to the right.

3. What is the field strength 3 cm from a wire carrying 6.0 amperes?

B = k(i/r) B = (2 * 10-7 web/amp-m)[(6.0 amp)/(0.03 m)] 4.0 * 10-5 web/m2

4. A charge q travels parallel to a wire with a velocity v. The wire, meanwhile, carries a current i. What force is exerted on the charge?

F = qvB B = k(i/r) F = qvk(i/r) F = (qvki)/(r)

5. A wire of length L lies at a distance r from an infinite wire carrying a current i. When the short wire is moved away at a velocity v, what current is generated in the resistor?

i = E/R i = (LvB)/(R) i = [Lvk(i/r)]/(R) i = (Lvki)/(Rr)

6. 8 cm from a long wire carrying 4.4 amps is a little circle of wire of 3 mm radius. What is the flux through the circle?

¤ = A * B ¤ = [(3.14)r12][K(I/r2)] ¤ = (3.14)(.003)2(2 * 10-7)(44/.08) ¤ = 3.11 * 10-10 web

7. Two wires, 15 cm apart, carry currents of 2 A and 3 A, respectively. Both currents are coming out of the page.
a) What is the strength of the field 10 cm to the right of the 2 A wire?
b) What is the field strength 10 cm to the left of the 2 A wire?

a) B1 = k(i1/r1) B1 = (2 * 10-7)(3/.05) B1 = 12 * 10-6 web/m2 B2 = k(i2/r2) B2 = (2 * 10-7)(2/.10) B2 = 4 * 10-6 web/m2 B = B1 - B2 B = 8.0 * 10-6 web/m2

b) B1 = k(i1/r1) B1 = (2 * 10-7)(2/.10) B1 = 4.0 * 10-6 web/m2 B2 = k(i2/r2) B2 = (2 * 10-7)(3/.25) B2 = 2.4 * 10-6 web/m2 B = B1 + B2 B = 6.4 * 10-6 web/m2

8. Two wires, carrying 4 A and 2 A, are crossed as shown. Find the magnetic field strength at points a and b.

a) B1 = k(i/r) B1 = (2 * 10-7)(4/.03) B1 = 2.67 * 10-5 out of screen B2 = k(i/r) B2 = (2 * 10-7)(2/.02) B2 = 2.0 * 10-5 into the screen B = B1 - B2 B = 6.7 * 10-6 web/m2 out of screen

b) B1 = k(i/r) B1 = (2 * 10-7)(2/.05) B1 = 8 * 10-6 out of screen B2 = k(i/r) B2 = (2 * 10-7)(4/.04) B2 = 20 * 10-6 out of screen B = B1 + B2 B = 28 * 10-6 web/m2 out of screen

9. A 17 cm length of wire masses 2.0 gm. It has a resistance of 0.8 ohms and is connected to battery E. How high a potential is needed to levitate the wire 1 cm over another wire carrying a 9 A current?

The 9 A wire generates a magnetic field: B = (2 * 10-7)(9/.01) B = 1.8 * 10-4 The wire responds to the field: mg = liB mg = l(E/R)B E = (mgR)/(lB) E = [(.002)(9.8)(.8)]/[(.17)(1.8 * 10-4)] E = 5.12 * 103 V

10. An electron is fired at 7 x 104 m/sec parallel to a wire carrying a 3 A current. If the electron travels 5 cm from the wire, what force acts on it?

F = qvB F = qvk(i/r) F = (1.6 * 10-19)(7 * 104)(2 * 10-7)(3/.05) F = 1.34 * 10-19 nt

11. Two wires lie parallel to each other and carry currents in the same direction. What is the direction of force wire a exerts on b? .......b exerts on a?

In the region of b, current ia generates a magnetic field out of the screen. Using F = liB and your left (negative) hand, it can be shown that b is forced up. ib generates a field into the screen, so wire a is forced down. This makes sense, as the force on a should be equal and opposite to the force on b.

12. Again two wires lie parallel, 6 cm apart, carrying currents of 3 A and 4 A in the same direction. What is the force per unit length that the 4 A wire exerts on the 3 A wire?

The 4 A wire generates a field: B = k(i/r) B = (2 * 10-7)(4/.06) B = 1.3 * 10-5 web/m2 F = liB Force/Length = F/l F/l = iB F/l = (3)(1.33 * 10-5) F/l = 4.0 * 10-5 nt/m

13. At a distance R from a wire carrying a current i is a small loop of wire of radius r. It, in turn, is connected through a resistor to two rails a distance L apart. A wire of mass m rests in field B. When the small loop is rotated 180° in time t, what is the acceleration of the wire?

Potential generated in the loop: E = (Ǥ)/(Çt) E = [ [k(i/r)][(3.14)r2] ]/(t/2) E = [2ki(3.14)r2]/(Rt) Electron flow in the bar generates a force: F = liB F = L(E/R)B F = [L2ki(3.14)r2B]/[R'Rt] a = F/m a = [2(3.14)kir2LB]/[mR'Rt] note: t/2 = time to flip 90°

14. A wire carries a current of 4.2 A. 13.0 cm away is a small rectangle of wire measuring 0.4 * 0.6 cm. The current in the main wire is reduced to 1.4 A in 0.03 seconds. What potential is generated in the rectangle?

E = (Ǥ)/(Çt) A = (ÇB)/(Çt) But B changes from (2 * 10-7)(4.2/.13) to (2 * 10-7)(1.4/.13) ÇB = (2 * 10-7)(4.2/.13) - (2 * 10-7)(1.4/.13) ÇB = 4.31 * 10-6 web/m2 E = A[(ÇB)/(Çt)] E = (.006)(.004)[(4.31 * 1065)/(.03)] E = 3.45 * 10-9 V

15. A loop of wire carries a clockwise current as shown. What is the direction of the magnetic field within the loop?

Grab the loop at the top with your negative hand. Note that your fingers come out through the inside of the loop. This will happen at any point you grab the loop.

When magnetic fields are turned on and off, their build-up and dispersion causes useful effects for electronic applications. First let us look at how a magnetic field builds up and decays.

a B-field increasing in strength:	IMAGE	IMAGE
a B-field decreasing in strength:	IMAGE	IMAGE

Consider a loop of wire in an increasing B-field.

Note that there is a flow of field lines into the region as the field builds, and out as the field disperses. We may determine the effect of this changing field by noting the relative motion of the electrons in a wire and the B-field. An electron on the top of the loop seems to move upward relative to the in flowing field lines, and so is forced to the right. Similar reasoning around the loop yields a clockwise movement.

Watch what happens when the clockwise electron flow is started up in the wire:

A magnetic field coming out of the screen is generated in the loop. Note that in this case the result of an induced current opposes what we are trying to do. That is, we are building up the B-field which results in an opposite field resisting the change in B.

The same thing will happen if we use a battery to generate a current in the wire. A B-field will be generated which will induce a current opposite to the one the battery generates.

Again, those pesky magnetic fields oppose our every action. This turns out to be a general principle.

Loops of wire in circuits are called inductors. Together with capacitors and resistors, they form the basic circuit elements.

16. An infinite wire carries a 1.6 amp current. Near it is a rectangular loop positioned as in the diagram, and carrying a 2.2 amp current.
a) What is the force on the segment of the loop farthest from the wire?
b) What is the force on the segment nearest the wire?
c) What is the net force on the loop? (Convince yourself that the forces on the two sides of the loop balance each other.)

a) Fa = liB Fa = lik(i/r) Fa = (.12)(2.2)(2 * 10-7)(1.6/.09) Fa = 9.37 * 10-7 nt down

b) Fb = liB Fb = lik(i/r) Fb = (.12)(2.2)(2 * 10-7)(1.6/.04) Fb = 21.1 * 10-7 nt up

c) F = Fb - Fa F = 11.7 * 10-7 nt up

17. A loop of wire is located near a long wire. As the current in the long wire increases, what direction will current be generated in the loop?

The current generates a field into the loop. As it builds in strength, more field lines are squeezed into the circle. This has the same effect as the loop expanding outward, which will force negative electrons in a clockwise direction.

18. A wire of length l is moved at velocity v a distance r from the long wire. This accelerates q through the gun into field B. Find the radius of its orbit.

First, the velocity of the charge: qV = (1/2)mv2 V = lvB V = lvk(i/r) vo2 = 2qV/m vo2 = (2qlvki)/(mr) qvoB = mvo2/ro ro = (mvo)/(qB) ro = m/(qB) ro = [(2qlvki)/(mr)]1/2 ro = (1/B)[(2mlvki)/(qr)]1/2

19. A coil of N turns of wire with radius r is held at distance R from a long current-carrying wire. It is rotated at a rate of n rev/sec and is connected to the resistors shown. Find the current in R₁.

The potential generated in the loop: E = N[(Ǥ)/(Çt)] E = N[ [(3.14)r2k(i/r)]/t ] But: t = (1)/(4n) E = [4nN(3.14)r2ki]/R But: i1 = E/RT i1 = [4nN(3.14)r2k(i/r)]/[ R1 + 1/[(1/R2) + (1/R3)] ] i1 = [4nN(3.14)r2ki]/[R(R1 + (R2R3)/(R2 + R3)]

20. Over the 16 cm length of the wires, what force exists?

F = liB F = li3k(i5/r) note: i3 and i5 are the currents in the wires iT = E/R iT = (30)/[2 + (1)/[(1/3) + (1/5)]] iT = 7.74 A i5 = (5)/(3 + 5)iT i5 = (5/8) * 7.74 i5 = 4.84 A i5 = iT - i3 i5 = 7.74 - 4.84 i5 = 2.90 A F = (lki35)/r F = [(.16)(2 * 10-7)(4.84)(2.90)]/(.03) F = 4.34 * 10-5 nt