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Unformatted text preview: LAST NAME: STUDENT NR: PHYS 1010 6.0: CLASS TEST 5 Time: 50 minutes; Calculators & formulae provided at the end = only aid; Total = 20 points. 1) [5] Given the following circuit diagram which contains an ideal battery, and three resistors, R1 , R2 , R3 . Complete the diagram by indicating how to measure: the current I2 through resistor R2 ; the voltage drop ∆V3 across resistor R3 . Then calculate the currents I1 , I2 and I3 , and the power dissipated by R3 . 1 2) [5] The diagram shows a nonideal battery (with internal resistance Rint ), and a simple resistorcapacitor network. Assume that just before the switch is set from charging to discharging (at time t = 0) the capacitors are fully charged. Give a formula for the discharge current (without derivation, use the formula sheet which gives a generic expression). Calculate the time constant, and graph the current as a function of time (properly marked current and time axis is required!) 2 3) [5] The ﬁgure shows a straight wire segment, and then a loop. The same current passes through. Three locations are marked, A, B are far away from the loop, so its contribution can be ignored. Location C is at the centre of the loop. Use the formulae provided to calculate the magnetic ﬁelds at A, B, C in the paper plane. A current of 1.5 A ﬂows through the wire. Be careful with location C , there are two contributions. The ﬁelds are to be speciﬁed by magnitude and direction or by listing the appropriate component with sign! 3 4) [5] The particle in the ﬁgure has a negative charge, and its velocity vector lies in the x − y plane and makes an angle of 75◦ with the y axis. A magnetic ﬁeld is along the +x direction. What is the direction of the magnetic force on the particle? Now you are told that the ﬁeld has a strength of 1.5 T, and that the particle speed is v = 500 m/s. Calculate the magnetic force. 4 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) = antiderivative = indeﬁnite integral area under the curve f (t) between limits t1 and t2 : F (t2 ) − F (t1 ) x2 + px + q = 0 factored by: x1,2 = − p ± p4 − q 2 uniform circular m. (t) = R(cos ω t ˆ + sin ω t ˆ); (t) = r i jv exp = exp; sin = cos; cos = − sin. m = Fnet ; a FG =
Gm1 m2 ; r2 d [f (g (x))] dx df dt df dt = 1 F (t) = f (t) dt = t2 2 +C 2 = 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 ). Fd ∼ − ; linear: Fd = dv ; quadratic: Fd = 0.5ρAv 2 ; A = cross sectional area v W = F ∆x = F (∆r) cos θ. W = area under Fx (x). P E H = k (∆x)2 ; P E g = mg ∆y . 2 avg ∆p = J = F (t)dt; ∆px = Jx = area under Fx (t) = Fx ∆t ; p = m ; K = m v 2 v 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 me = 9.11 × 10−31 kg mp = 1.67 × 10−27 kg e = 1.60 × 10−19 C FC =
Kq1 q2 ˆ r r2 K= 1 4π0 FE = q E Eline = 2K λ r = 2K Q Lr Eplane = η  20 =
2 Q 2A0 Ecap = mv 2 + Uel (s) 2 = 2 mv0 + Uel (s0 ), 2 C C Q = C ∆VC farad = F = V C = 0dA 0 = 8.85 × 10−12 Nm2 − − − parallel C1 , C2 : Ceq = C1 + C2 series C1 , C2 : Ceq1 = C1 1 + C2 1 ∆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 3 4π r 4π r 2π d 4π A (U ≡ P E el ) Uel = qEx for E = −E ˆ i Vel = Uel /q = 9.0 × 109 Nm C2
Q , pos 0 A 2 → neg Ex = − dVel dx tesla = T = 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 force betw. parallel wires: F2wires = µ0 LI1 I2 2π d τ torque on mag dipole: µ in B : = µ × B 5 ...
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This note was uploaded on 09/28/2010 for the course MATH 1025 taught by Professor Tammie during the Fall '10 term at York University.
 Fall '10
 Tammie
 Linear Algebra, Algebra

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