Electricity and Magnetism Force on Parallel Wires Lana Sheridan De Anza College Mar 2, 2018
Last time Gauss s Law for magnetic fields Ampère s Law magnetic field around a straight wire solenoids
Overview magnetic field inside a solenoid forces between current-carrying wires
Solenoids Additional examples, video solenoid A helical coil of tightly wound wire that can carry a current. turn i Fig. 29-16 A solenoid carrying current i. i 29-5 Solenoi Magnetic Field o We now turn our useful. It concerns wound helical coil that the length of t Figure 29-17 sh The solenoid s ma A single complete loop of wire in a solenoid. This solenoid has 10 turns, means it has 10 complete loops.
apply Ampere s law, Magnetic Field of an ideal solenoid In an ideal solenoid B : ds : (with 0 i enc, infinite length) the field(29-21) outside is small (and perpendicular to the Amp. loop) and inside is uniform. enoid (Similar of Fig. 29-19, to a where capacitor!) B : is uniform within the solenoid and using the rectangular Amperian loop abcda. We write B : ds : as Can use an Amperian loop to find the B-field inside: d h c i B ds = µ 0 I enc B a b plication of Ampere s law to a section of a long ideal solenoid carrying mperian loop is the rectangle abcda.
apply Ampere s law, Magnetic Field of an ideal solenoid In an ideal solenoid B : ds : (with 0 i enc, infinite length) the field(29-21) outside is small (and perpendicular to the Amp. loop) and inside is uniform. enoid (Similar of Fig. 29-19, to a where capacitor!) B : is uniform within the solenoid and using the rectangular Amperian loop abcda. We write B : ds : as Can use an Amperian loop to find the B-field inside: d h c i B ds = µ 0 I enc B a b plication Here, of Ampere s supposelaw there to a section are n of turns a long per ideal unit solenoid lengthcarrying the solenoid, mperian then loop Iis enc the = rectangle Inh abcda. Bh = µ 0 Inh Inside an ideal solenoid: B = µ 0 In
Two Wires carrying Current With two long, straight current carrying wires, each creates its own magnetic field: B = µ 0I 2πa The result is that the wires interact, much like two bar magnets producing magnetic fields would.
Forces on parallel wires Currents in opposite directions repel, currents in the same direction attract. 1 Figure from salisbury.edu.
Forces on parallel wires 1 Figure from Stonebrook Physics ic.sunysb.edu.
Forces on parallel wires To find the magnitude of the force, we need to recall how force relates to the magnetic field for a wire. The force on a long, straight current carrying wire is: F = IL B
Forces on parallel wires 770 CHAPTER 29 MAGNETIC FIELDS DUE TO C Suppose that wire a produces a field at b: B a = µ 0I a 2πd i a The force on wire b is: The field due to a at the position of b creates a force on b. i b Fig. 29-9 Two parallel wires carrying currents in the same direction attract each other. B : a is the magnetic field at wire b produced by the current in wire a. F : ba is the resulting force acting on wire b because it carries current in. B : a F ba a L L d b B a (due to i a ) F b = I b L B a = I b L µ 0I a 2πd sin(90 ) [toward a] F B = µ 0I a I b 2πd L 29-3 Force Two long parall shows two such Let us analyze th We seek firs current produce causes the force direction of the wire b is, from E The curled stra down, as Fig. 29- Now that w Equation 28-26
Forces on parallel wires It is a bit more intuitive to think about the force per unit length on the wires (since longer wires will experience larger forces). The force per unit length on a wire due to another parallel wire at a distance d: F B L = µ 0I 1 I 2 2πd
10 6 g,and then launches it with a speed of 10km/s,all within 1ms.S rail Question guns may be used to launch materials into space from mining opera e Moon or an asteroid. The figure here shows three long, straight, parallel, equally spaced HECKPOINT 1 wires with identical currents either into or out of the page. Rank e figure thehere wires shows according three to long, thestraight, magnitude parallel, of theequally force on spaced each wires due towith iden rrents the either currents into or in out the of other the two page.rank wires, greatest the wires first. according to the magnitud e force on each due to the currents in the other two wires, greatest first. a b c A a, b, c 4 Ampere s Law B b, c, a an find C c, the b, net a electric field due to any distribution of charges by first w ifferential electric field de : due to a charge element and then summin ributions of de : from all the elements. However, if the distribution is co d, we may have to use a computer. Recall, however, that if the distrib 1 Halliday, Resnick, Walker, pg 771.
10 6 g,and then launches it with a speed of 10km/s,all within 1ms.S rail Question guns may be used to launch materials into space from mining opera e Moon or an asteroid. The figure here shows three long, straight, parallel, equally spaced HECKPOINT 1 wires with identical currents either into or out of the page. Rank e figure thehere wires shows according three to long, thestraight, magnitude parallel, of theequally force on spaced each wires due towith iden rrents the either currents into or in out the of other the two page.rank wires, greatest the wires first. according to the magnitud e force on each due to the currents in the other two wires, greatest first. a b c A a, b, c 4 Ampere s Law B b, c, a an find C c, the b, net a electric field due to any distribution of charges by first w ifferential electric field de : due to a charge element and then summin ributions of de : from all the elements. However, if the distribution is co d, we may have to use a computer. Recall, however, that if the distrib 1 Halliday, Resnick, Walker, pg 771.
Definition of the Ampère (Amp) This relation: F B L = µ 0I 1 I 2 2πd gives us the formal definition of the Ampère. Ampère Unit Two long parallel wires separated by 1 m are said to each carry 1 A of current when the force per unit length on each wire is 2 10 7 N/m.
Summary Solenoids and Ampère s Law Forces on parallel wires 3rd Test Friday, Mar 9. Homework Collected homework 3, posted online, due on Monday. Serway & Jewett: PREVIOUS: Ch 30, Problems: 21, 25, 31, 33, 34, 47 NEW: Ch 30, Problems: 41, 45