Wkb Chap 12 Magnetism, what causes it

29.3 Torque on a Current Carrying Loop in Magnetic Field B:

t = BIAsin q, where q is the angle between B and normal to loop.

Ampere's Law: äBds = µ_oI, where I is the total current passing thru the loop.

B_lineds = B_line ds cos 0, so äBds = B_lineS = 2πr = µ_oI

-Uuseful to obtain B a fixed distance from a current configuration

- Like Gauss' Law for E, most useful for symmetric configurations

- Example: Find B a distance r from the center of a toriodial coil with N loops.

Note: Total current passing thru the constructed loop (dotted circle) of radius r is NI, B & ds are parallel, so äBds = Nµ_oI, B is constant, so äBds = B äds = B2πr =Nµ_oI, thus, B_toroid = (Nµ_oI) / (2πr)

Question: What would B_loop be if the loop were drawn completely inside the toroid?

What would B_loop be if the loop were drawn to completely enclose the entire toroid?

So, where would one use a toroid?

30.4 Magnetic Field inside a Solenoid

By Ampere's Law, B_outside = 0, (net I thru any loop surrounding the entire solenoid = 0)

Bds = 0 on sides 2 & 4 because q = 90^o, and on side 3 because B = 0 outside the toroid

so, äBds = _₁B ds + _₂Bds + _₃Bds + _₄ Bds = BL + 0+0+0 = Nµ_oI, so B = (Nµ_oI)/L = µ_onI

Note however that this assumes that B is constant, so this formula only works in the middle of a long solenoid. For short solenoids,

B(x) = (µ_oNI)/(2L)(sin q₂ - sin q₁)

Do #

Note that at the end of a solenoid, q₁ = 0 & q₂ ≈ 90⁰, so

B_end = , and since at the midpoint of the solenoid q₁ = -q₂ , B_midpoint =

Magnetic Flux F_m = _BdA = BAcos q, (if B is constant). It has units of Webbers W_b = Tm².

Gauss' Law for Magnetism: äBdA = 0, (compare to Gauss' Law for E, F_E = äEdA = Q_inside / e_o)

And where, you might ask, would one use a solenoid? How about electric guitars?

Ampere's Law revisited. Why does a magnetic field get produced between the plates of a capacitor when I changes direction, (when it shoudn't, since no current flows between the plates, according to Ampere's Law, B = µ_oI = 0)? Well, because C = e_oA/d, V_C = q/c = Ed, so q_plate = e_oAE, so I_btwn
plates = dq_plate/dt = e_oAdE/dt = e_o dF_E/dt !! This is called displacement current I_d, and the now complete Ampere's Law is

äBds = µ_o(I + I_d), and you won't be tested over it, but you heard it first here in Phy 214 at CACC.

Summary

Biot-Savart Law - gives the magnetic field at a point P due to a line of current I.

dB = [--- Unable To Translate Text Box ---]

µ_o = 4πx10^-7 Tm/A

Ampere's Law: äBds = µ_oI, where I is the total current passing thru the loop.

Magnetic Flux F_B = _BdA , and

Gauss' Law for Magnetism: äBdA= 0

B_{long line} = µ_oI/(2πa)

B_{short line} = [--- Unable To Translate Text Box ---]

B_loop =

B_{toroidial coil} = (µ_oNI)/(2πr) where r is the mean radius of the toroid, and N is the number of loops.

B_sole(x) = (µ_oNI)/(2L)(sin q₂ - sin q₁)

B_midpoint =

B_{end of sole} =

B_{very long} = (µ_oNI)/(L)= µ_onI

Motors and Generators

An AC generator is N loops of wire rotating in a magnetic field with angular velocity w.

Then F_m = NBA cos (wt +q_o) q = wt + q_o, remember?

and x_generated = - NAB (d/dt) cos(wt + q_o) = NABwsin (wt + q_o) What does x_max = ? Current is thus produced and I = x /R

Now an AC motor is a obtained by passing current thru N loops of wire in a magnetic field. The torque produced causes the motor to spin. Now since this torque is always perpendicular to the loop, no work is done and once started, the unopposed motor should spin forever... but it doesn't. Why not? Because of Lenz's Law. The rotating loop tries to set up a current going in the opposite direction according to the formula for x for AC generators. So the useful work in an electric motor is driven by the current I = x_net/R = (x_applied - x_back)/R

Assign
Faraday's Law of Induction

Faraday's Law of Induction - Electric fields are induced by a changing Magnetic flux.

- It says that a current (and hence an emf) is setup in a closed conducting loop if the magnetic flux thru that loop is changing with time, namely,

x = -dF_m/dt = -d(BAcos q)/dt, or, if have N loops, x = -(N dF_m/dt)

Note 1: x is induced if B, A, or q changes in time!

Note 2: The negative sign is a consequence of Lenz's Law, which says the current induced will be in a direction so that the magnetic field it creates will try to prevent a change in F_m.

do , Assign

Example 1: Suppose you have a conductor moving in a magnetic field. The charges in the conductor feel F_B = qvB, and the electrons will continue to move until the potential difference between the ends of the conductor is strong enough to produce an electric field that equalizes the magnetic force. i.e. F_E = qE = qvB, or E = vB. Note: x = Blv

do , Assign

Example 2: Find the power delivered by the applied force F_app in the diagram to the resistor R.

F_m = BA = Blx, x = -dF_m/dt = Blv = IR, so I = Blv/R . Now the force on the moving conductor (line of current), is F_m = BIL. Constant velocity means F_m = F_app, so the power delivered by F_app to R is P = F_appv = F_mv = (BIL)v = B (Blv/R) lv = B²L²v²/R

Do #, Assign #

Lenz's Law: Remember, motion toward the loop implies B_increasing and motion away from the loop implies B_decreacing

Examples: Do all possible combinations with the loop and a magnetic.

Induced emfs and Electric fields

do . Note: x = n Eds = -dF_m/dt = d(BAcos q)/dt = (d/dt)(BA₁cosq) + (d/dt)(BA₂ cos q) = µ_oNA/L(dI/dt) + 0

Assign

Inductance

What happens when the switch is thrown? The current in the wire creates a magnetic field B_inc into the page inside the loop. This ∆F_m creates x_ind that tries to send current back the other way. x_ind = -N∆F_m/dt = -N∆(BAcos q)/dt. But in any fixed circuit, q = 0, A is fixed, and B is some constant multiple of I, hence, -N∆F_m/dt = -LdI/dt, where L is this constant, is called Inductance, has units of Vs/A = kg m²/C² = Henry's (H). Integrating both sides with respect to t yields NF_m = LI, or L = NF_m/I (official definition of Inductance), so x_ind = x_L = x_back = -LdI/dt, or L = -x_ind /(dI/dt). In particular, for a solenoid of length L, L_sole = N²µ_oA/ L = constant ! (like capacitance).

Do 2 & 10, Assign 3,7,9

32.2 RL Circuits How does the current behave? By the loop rule, x - IR -LdI/dt = 0, a differential equation whose solution produces:

Charging: I = (x/R)(1 - e^-Rt/L) = (x/R)(1 - e^-t/^t), where t = L/R called the RL time constant.

Discharging: I = (x/R)e^-t/^t = I_oe^-t/^t. Note: t is the time it takes I to reach 1-e^-1 = .63 of its maximum value.

Do , Assign

32.3 Energy stored in an inductor. U_m = LI², U_{m sole} = B² AL/(2µ_o)

Do 14, 18,22 , Assign .

Mutual Inductance M_{2 due to 1} N₂F_{2 due to 1}/I₁ , x₂ = -N₂ dF_{2 due to 1} / dt = -M_{2 due to 1}(dI/dt)

Do 30,36 , Assign

32.5 LC Circuits U = U_c + U_L = Q²/(2C) + LI²/2

When the switch is thrown, with the capacitor charged up, we get the loop equation Ld²Q/dt² + Q/C = 0, which is the dif. eq. of motion for Simple Harmonic Motion (SHM), with angular frequency w = , moreover, at time t,

Q = Q_max cos (wt + d), and I = -Q_maxwsin(wt + d), do46, 48, Assign 47,49,53

32.6 RLC Circuits Is like a damped harmonic oscillator

Q = Q_me^{-Rt
/ (2L)}cos(w_dt), with w_d = R_c=(4L/C)^.5

Do # 54, Assign # 55
AC Circuits

1 Pure Resistive Circuits current does not affect v_R , so i_R and v_R are in phase

v = V_max sin wt, w = 2πf = 2π/T

v_R = I_max R sin wt

i_R = v/R = (V_max/R)sin wt = I_maxsin wt

P = i²R

Root mean square voltage and current: a kind of average value v_rms = v /√2, i_rms = i /√2

2 Pure Inductive Circuit: current lags voltage 90^o The voltage dropped across an inductor is the induced x_back which is created by changing magnetic flux, and is greatest when the switch is first thrown, hence, current lags voltage 90^o.

v + v_L = 0 = v - Ldi/dt, from whence, v = v_max sin wt, and i_L = (V_max/wL)sin (wt - π/2)

I_max = V_max/(wL) = V_max/c_L , c_L = wL = Inductive Reactance and has units of Ω !

v_L = V_max sin wt = I_max c_L sin wt

3 Pure Capacitive Circuit: current leads voltage by 90^o Potential difference across the capacitor is maximum after the current dies, hence current leads voltage by 90^o in a capacitive circuit.

v - v_C = o, from whence v = v_c sin wt = q/C, and q = CV_max sin wt, so i = dq/dt = wCV_max cos wt = wCV_max sin(wt - π/2)

I_max = wCV_max = V_max /c_C . c_C = 1/(wC) = 1/(2πfC), is called capacitive reactance and has units of Ω. v_c = I_max c_C sin wt

4 RLC series Circuits

v = V_max sin wt = v_R + v_C + v_L

i = I_max sin(wt - f) and is the same in all elements in the circuit

f =

V_m = = I_max = I_max Z

Resonant (angular) frequency w_o occurs where c_L - c_C = 0, or when wL = 1/(wC), so w_o = 1/√LC

33.5 Power in AC Circuits

P_avg = .5 I_max V_maxcos f (cos f is called the power factor)

= I_rms V_rms cos f

= I_rms²R

5 Transformers: N₁V₁ = N₂V₂, and I₁ V₁ = I₂ V₂

_Assign: