PHYS152 Lecture 8. Eunil Won Korea University. Ch 30 Magnetic Fields Due to Currents. Fundamentals of Physics by Eunil Won, Korea University

Transcription

1 PHYS152 Lecture 8 Ch 3 Magnetic Fields Due to Currents Eunil Won Korea University

2 Calculating the Magnetic Field Due to a Current Recall that we had the formula for the electrostatic force: d E = 1 ɛ dq r 3 r One way to produce a magnetic field is to have a current segment: db = µ i ds sin θ µ : permeability constant, whose value is defined to be exactly µ = 1 7 T m/a 1.26 x 6 T m/a r 2 In vector form, d B = µ i d s r r 3 r (Bio-Savart law)

3 Magnetic Field Due to a Current in a Long Straight Wire Noting that Right hand rule: Grasp the element in your right hand with your extended thumb pointing in the direction of the current. Your fingers will then naturally curl around in the direction of the magnetic field lines due to that element From B = 2 db = µ i ds sin θ r 2 db = µ i 2π sin θ r 2 r = s 2 + R 2, sin θ = sin π θ = B = µ i 2π R ds (s 2 + R 2 ) 3/2 Let s = R tan θ, ds = R sec 2 θdθ ds R s2 + R 2 s 2 + R 2 = R 2 (1 + tan 2 θ) = R 2 sec 2 θ sec θ = 1 cos θ Therefore, ds (s 2 + R 2 ) 3/2 = So, R sec 2 θ dθ R 3 sec 3 θ = B = µ i 2π R 1 R 2 cos θ dθ = 1 R 2 sin θ + C = 1 R 2 s R2 + s 2 + C ( 1 R 2 ) s R2 + s 2 = µ i 2πR

4 Magnetic Field Due to a Current in a Circular Arc of Wire The angle between the displacement and the current is always 9 o db = µ i ds sin 9 R 2 = µ i ds R 2 The total field becomes B = db = φ µ ir dφ R 2 = µ i R φ dφ Integrating, we find that B = µ iφ R Now, at the center of a full circle of current: B = µ i(2π) R = µ i 2R

5 Force Between Two Parallel Currents We seek first the force on wire b due to the current in wire a The magnitude of B at wire b due to the wire a is: B a = µ i a 2πd F ba : force on a length L of wire b due to the external magnetic field above From the previous chapter, we know that: F ba = i b L Ba F ba = i b LB a sin 9 = µ Li a i b 2πd and the direction of the force is toward the wire a Thus, parallel currents attracts, and antiparallel currents repel

6 Ampere s Law Ampere s law states that B d s = µ i enc d s i enc a closed loop integral the net current encircled by the loop i enc = i 1 i 2 B cos θ ds = µ (i 1 i 2 ) (A right-hand rule for Ampere s law, to determine the signs for currents encircled by an Amperian loop)

7 The Magnetic Field Due to a Long Straight Wire with Current Outside From Ampere s law: B d s = B cos θ ds = B B d s = µ i enc ds = B(2πr) = µ i B = µ i 2πr (It agrees with the result with using the law of Biot-savart) Inside From Ampere s law: B d s = B ds = B(2πr) B(2πr) = µ i r2 R 2 B = µ i 2πR 2 r

8 Solenoid Solenoid: a long, tightly wound helical coil of wire (the picture on the left shows a stretched-out solenoid) Ampere s law states that B d s = = B d s = µ i enc b a d c B d s + B d s + c b a d B d s B d s (Non-zero integral is from a to b) i enc = i(nh) n: number of turns per unit length of the solenoid Bh = µ inh B = µ in (ideal solenoid)

9 Toroid Toroid: a solenoid bent into the shape of a hollow doughnut Take Amperian loop as a circle of the radius r: B(2πr) = µ in N : the total number of turns B = µ in 2π 1 r (note: B is not constant anymore)

10 A Current-Carrying Coil as a Magnetic Dipole A coil behaves as a magnetic dipole From the previous chapter, we know that (a torque on a coil in the presence of magnetic field B) τ = µ B Magnetic field due to a coil at a point z on the axis of the coil? (point P in the figure left) B = db Biot-Savert law tells us that db = µ i ds sin 9 r 2 (the net contribution of the perpendicular components is zero) db = db cos α so db = µ i cos α ds r 2 r = R 2 + z 2, cos α = db = µ i B = db = R R2 + z 2 R 1 R2 + z 2 (R 2 + z 2 ) ds = µ i ds (R 2 + z 2 ) 3/2 µ i (R 2 + z 2 ) 3/2 2πR = µ ir 2 (R 2 + z 2 ) 3/2

11 Summary The Biot-Savart Law d B = µ i d s r r 3 r Magnetic Field of a Long Straight Wire B = µ i 2πR Ampere s Law B d s = µ i enc Solenoid B = µ in