test6_10Soln - LAST NAME STUDENT NR PHYS 1010 6.0 CLASS...

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Unformatted text preview: LAST NAME: STUDENT NR: PHYS 1010 6.0: CLASS TEST 6 Time: 50 minutes; Calculators & formulae provided at the end = only aid; Total = 20 points. 1) A current loop with a resistance R = 250 Ω and an area A = 0.4 m2 is oriented perpendicular to a magnetic field that varies in time according to B (t) = 0.5t(1 − t) (t and B are in SI units). What is the current induced in the loop at t = 0 s, at t = 0.5 s, and at t = 1 s? 1 2) A circular loop of wire lies in the x − y plane so that the axis of the loop lies along z . A homogeneous (spatially uniform) magnetic field B (t) is anti-parallel to the z-axis. (Bz (t) < 0, Bx = By = 0). If B (t) is decreasing with time, what is the direction of the induced current when viewed from above? Start with a drawing, and explain your steps and reasoning! 3) Two inductors L1 and L2 are connected in series to form an equivalent inductor Leq . Together they are connected to a battery of voltage ∆VB via a switch. Draw a circuit diagram depicting the situation. Formulate Kirchhoff ’s loop law for the circuit. Derive how Leq depends on L1 and L2 . 2 ... 4a) Consider a simple LC circuit (L and C in parallel) with values C = 1500 pF and L = 20 mH. At t = 0 the current is I = 45 mA and the charge on C is zero. Calculate the energy stored in the inductor at this time. At which time t1 will the current be zero (first occurrence after t = 0). What is the charge in C at this time? Provide an accurate sketch of the charge as a function of time for two complete cycles. 3 4b) The LC circuit of question 4 has the inductor replaced by a L = 20 µH coil. What is the wavelength of the radiowaves that this circuit will catch on resonance? FORMULA SHEET ￿ v (tf ) = v (ti ) + ti a(t) dt s(tf ) = s(ti ) + ttif v (t) dt vf = vi + a∆t sf = si + vi ∆t + 1 a∆t2 vf2 = vi2 + 2a∆s 2 ￿ tf g = 9.8 m/s2 f (t) = t f (t) = a = 0 F (t) = f (t) dt = at + C F (t) = anti-derivative = indefinite integral area under the curve f (t) between limits t1 and t2 : F (t2 ) − F (t1 ) −q uniform circular m. ￿(t) = R(cos ω t ˆ + sin ω t ˆ); ￿ (t) = r i jv x + px + q = 0 factored by: x1,2 = ￿ ￿ ￿ 2 df dt df dt = 1 F (t) = ￿ ￿ f (t) dt = t2 2 +C −p 2 ± ￿ p2 4 exp = exp; sin = cos; cos = − sin. m￿ = Fnet ; a￿ FG = Gm1 m2 ; r2 d [f (g (x))] dx = d￿ r dt df dg ; dg dx = ...; ￿ (t) = a ￿ ￿ d￿ v dt = .... (f g ) = f g + f g ￿ 2 g= GME 2; RE a a in in fin fin ∆p1 + ∆p2 = 0 ; K1 + K2 = K1 + K2 for elastic collisions. ￿ CM = m1￿ 1 +m2￿ 2 ￿ ￿ a m1 +m2 ￿ 2 ˆ ￿ = ￿ × F ; τz = rF sin(α) for ￿, F in xy plane. I = i mi ri ; I αz = τz ; (k = rot. axis) τr￿ r￿ ￿r￿ ￿τ Krot = I ω 2 ; Lz = I ωz ; d Lz = τz ; L = ￿ × p; d L = ￿ 2 dt dt fs ≤ µs n; fk = µk n; fr = µr n; µr << µk < µs . FH = −k ∆x = −k (x − x0 ). 2 ￿ Fd ∼ −￿ ; linear: Fd = dv ; quadratic: Fd = 0.5ρAv ; A = cross sectional area v W = F ∆x = F (∆r) cos θ. W = area under Fx (x). PEH = k (∆x)2 ; PEg = mg ∆y . 2 ￿ = ￿ F (t)dt; ∆px = Jx = area under Fx (t) = F avg ∆t ; p = m￿ ; K = m v 2 ￿ ∆p = J ￿ ￿ v x 2 RE = 6370 km; G = 6.67 × 10−11 Nm2 ; ME = 6.0 × 1024 kg kg x(t) = A cos (ω t + φ); ω = 2π = 2π f ; vx (t) = ...; vmax = ... T −31 me = 9.11 × 10 kg mp = 1.67 × 10−27 kg e = 1.60 × 10−19 C ￿ FC = Kq1 q2 ˆ r r2 K= 1 4π￿0 ￿ ￿ FE = q E Eline = 2K |λ| r = 2K |Q| Lr Eplane = |η | 2￿0 = |Q| 2A￿0 ￿ Ecap = mv 2 + Uel (s) 2 = 2 mv0 + Uel (s0 ), 2 (U ≡ P E el ) ￿ Uel = qEx for E = −E ˆ i 4 Vel = Uel /q ￿ = 9.0 × 109 Nm C2 Q , pos ￿0 A 2 → neg Ex = − dVel dx ￿ C C Q = C ∆VC farad = F = V C = ￿0dA ￿0 = 8.85 × 10−12 Nm2 PEC = C ∆VC 2 2 −1 −1 −1 parallel C1 , C2 : Ceq = C1 + C2 series C1 , C2 : Ceq = C1 + C2 ￿ ￿ ￿ ∆Vloop = i ∆Vi = 0 Iin = Iout P = ∆V I watt = W = VA PR = ∆VR I = I 2 R V τ = RC Q(t) = Q0 e−t/τ I (t) = − dQ = ∆R 0 e−t/τ dt vr sr ￿ B = µ0 q￿ ×￿ = µ0 I ∆￿3×￿ Bwire = µ0 I (use RH rule) µ0 = 10−7 Tm tesla = T = 3 4π r 4π r 2π d 4π A 2 short coil, R >> L (N turns): Bcoil,centre = µ0 N I 2R solenoid, L >> R: N Am N Bsol,inside = µ0L I µ0 ￿ ￿ mag dipole: ￿ = (AI, from south to north) Bdip = 4π 2µ on axis, far away µ z3 ￿onq = q￿ × B force on current ⊥ to B : Fwire = ILB ￿ ￿ F v LI1 ￿τ ￿ ￿ force betw. parallel wires: F2wires = µ02πd I2 torque on mag dipole: µ in B : ￿ = µ × B ￿ ￿ bar (length L) moves w. ￿ ⊥ B gen. EMF: ε = vLB ; v ￿ ￿ ￿ ￿ ￿ ￿ ￿ ￿￿ ￿￿ ￿ ￿￿ Φm = A · B Φm = AB cos θ ε = ￿ dΦm ￿ = ￿B · dA + A · dB ￿ L= Φm I henry = H = L R Tm2 A series L and R: τ = λf = v w I (t) = I0 e−t/τ ; parallel L and C: ω = ￿ x A sin (2π ( λ T µ ￿￿ εcoil = L ￿ dI ￿ ￿ dt ￿ dt dt dt ∆VL = −L dI dt PEL = L I 2 2 − ￿ 1 LC t )+ T I (t) = ω Q0 sin ω t φ0 ) = A sin (kx − ω t + φ0 ) sinusoid travelling in pos dir’n: D(x, t) = transverse wave on a string: vw = where T is tension, µ = M/L ω = vw k vsound = 343m/s in air at T = 20◦ C in water: vsound = 1480m/s light in vac.: vw = c = 3.00 × 108 m/s visible: λ = 400nm (blue/UV); λ = 700nm (red/IR) refraction: nglass = 1.5; nwater = 1.333; light in medium: c/n; wavelength: λvac /n f f approaching source (speed vs ): f+ = 1−vs0/vw receding: f− = 1+vs0/vw sin (α ± β ) = sin α cos β ± cos α sin β sin α + sin β = 2 cos α−β sin α+β 2 2 L transverse standing wave, string length L: λn = 2n n = 1, 2, ... fn from λn fn = cw 5 ...
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