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Exercise 1 Exercise&1&& Solution Exercise&1&& II II aa bb A current I = 2.5 A is flowing from left to right through a straight aluminum wire A current I = 2.5 A is flowing from left to right through a straight aluminum wire with with aa 1/2) mm and length L=5.0 m, as shown above circular cross section of diameter D=(0.3 / π 1/2 circular cross section of diameter D=(0.3 / π ) mm and length L=5.0 m, as shown above -8 (not (not drawn drawn to to scale!). scale!).Aluminum Aluminum has has aa resistivity resistivity of of 2.7×10 2.7×10-8Ω Ωm m (at (at room room temperature). temperature). (a) (a) Find Find the the voltage voltage drop drop between between the the two two ends ends of of the the wire, wire, aa and and b, b, V = V − V . ab a b Vab = Va − Vb . Exercise&1&& Here, Here,V Vaa and and V Vbb are are the the electrical electrical potential potential values values at at the the left left end, end, a, a, and and the the right right end, end, b, b, respectively. Is V > 0 or V < 0 ? [Hint: By convention of current flow direction, ab ab respectively. Is Vab > 0 or Vab < 0 ? [Hint: By convention of current flow direction, the the electric electric current current in in aa wire wire or or other other resistor resistor always always flows flows from from high high to to low low electric electric potential.] potential.] I I (b) in Cross-‐sectional rea of of wheat ire: energy (b) Find Find the theaamount amount of heat energy generated generated in the the wire wire ifif the the current current flows flows for for 55 h. h. 2 1/2 -‐3 2 -‐8 A =Find π (D/2) = π [ strength, (0.3/(2 ×inside π )) ×wire, (10 m) ] = 2.25 × 10 m2 (c) (c) Find the the electric electric field field strength, E, E, inside the the wire, and and the the direction direction of of the the E-vector. E-vector. Does Resistance: Does EE point point from from left left to to right right or or from from right right to to left? left? [Hint: [Hint: the the electric electric field field vector vector always always points direction of electric potential.] R = ρinin Lthe the / A = (2.7 ×decreasing 10-‐8 Ω m ) (5.0 m) / (2.25 × 10-‐8 m2) = 6.0 Ω points direction of decreasing electric potential.] a b Ohm’s Law: A current I = 2.5 A is flowing from left to right through a straight aluminum wire with a |V R I =section (6.0 of Ωdiameter ) (2.5 AD=(0.3 ) = 15 / V ) mm and length L=5.0 m, as shown above ab | =cross circular π1/2 Sign of V(not : ab drawn to scale!). Aluminum has a resistivity of 2.7×10-8 Ω m (at room temperature). Vab = Va − Vb = +15V >0 (a) Find the voltage drop between the two ends of the wire, a and b, Reasoning: V = V − V . ab a b Current through resistor always flows from high to low potential; and I here flows Here, Va and Vb are the electrical potential values at the left end, a, and the right end, b, Is VV > 0 or V < 0 ? [Hint: By convention of current flow direction, the from a respectively. to b, hence > V . a ab b ab electric current in a wire or other resistor always flows from high to low electric potential.] (b) Find the amount of heat energy generated in the wire if the current flows for 5 h. (c) Find the electric field strength, E, inside the wire, and the direction of the E-vector. Joule heating Power = Rate of heat generation in wire: Does E point from left to right or from right to left? [Hint: the electric field vector always P = I V =direction (2.5 A) 15V) = 3electric 7.5 Wpotential.] points inab the of(decreasing Or use: P = R I2 = (6.0 Ω ) (2.5 A)2 = 37.5 W Note here that P is proportional to I2 for fixed resistance R. Heat energy generated during t=5h: H5h = P t = (37.5 W) (5 h) (3600s/h) = 675 kJ. Note here that H5h is also proportional to I2 for fixed resistance R and fixed time t. . Vab = Va − Vb . Here, Va and Vb are the electrical potential values at the left end, a, and the right end, b, respectively. Is Vab > 0 or Vab < 0 ? [Hint: By convention of current flow direction, the electric current in a wire or other resistor always flows from high to low electric potential.] (b) Find the amount of heat energy generated in the wire if the current flows for 5 h. (c) Find the electric field strength, E, inside the wire, and the direction of the E-vector. Does E point from left to right or from right to left? [Hint: the electric field vector always points in the direction of decreasing Exercise&1&(cont’d.)&& electric potential.] E = |Vab| / L = (15V)/(5.0m) = 3.0 V/m I I The E-‐vector always points in direction of decreasing potential, V. Since Va > Vb here [see part (a)], the E-‐vector points from a to b, i.e., E-‐vector points from left to right, a& b& inside the w ire. A current I = 2.5 A is flowing from left to right through a straight aluminum wire with a circular cross section of diameter D=(0.3 / π1/2) mm and length L=5.0 m, as shown above . (not drawn to scale!). Aluminum has a resistivity of 2.7×10-8 Ω m (at room temperature). I (d) The charge carriers in the wire are electrons and the current, I, is due to the average “drift” motion of these “conduction electrons” inside the wire (see/review FlipIt Physics). Exercise&1&(cont’d.)&& In which direction, on average, do these electrons move: left to right or right to left? [Hint: An electron has negative charge, q = −e < 0. Consider the direction of the electric force, F, acting on the electron, due to the electric field, E, found in part (c).] (e) Find the change in the electric potential energy, ΔU, incurred by a conduction electron, as it moves from one end of the wire to the other, due to the current flow. Does the Force oelectron n electron: gain potential energy (ΔU>0) or lose potential energy (ΔU<0)? I [Hint: difference, Vab, found in part (b); or calculate the work done on the F =Use qthe E potential = (-‐e) E electron by the electric force F , acting on the electron, due to the electric field E ]. a& So, the F-‐vector acting on each electron points in direction opposite to b&the E-‐vector, A current I = 2.5 A is flowing from left to right through a straight aluminum wire with a i.e., from right cross to left. Hence the D=(0.3 / π1/2) mm and length L=5.0 m, as shown above circular section of diameter (not drawn to scale!). Aluminum has a resistivity of 2.7×10-8 Ω m (at room temperature). electrons‘ drift motion is from right to left (d) The charge carriers in the wire are electrons and the current, I, is due to the average “drift” motion of these “conduction electrons” inside the wire (see/review FlipIt Physics). through he wire. Intwhich direction, on average, do these electrons move: left to right or right to left? [Hint: An electron has negative charge, q = −e < 0. Consider the direction of the electric force, F, acting on the electron, due to the electric field, E, found in part (c).] (e) Find the change in the electric potential energy, ΔU, incurred by a conduction electron, as it moves from one end of the wire to the other, due to the current flow. Does the electron gain potential energy (ΔU>0) or lose potential energy (ΔU<0)? [Hint: Use the potential difference, Vab, found in part (b); or calculate the work done on the electron by the electric force F , acting on the electron, due to the electric field E ]. Since the electron moves through the wire from right to left, i.e., from b to a, its change in potential energy is ΔU = Ua – Ub = q (Va – Vb) = q Vab = (-‐1.6×10-‐19 C) (+15V) = −2.4×10-‐18 J < 0 , using q=−e=−1.6×10-‐19C for the electron charge. The fact that ΔU<0 means that the electron loses potential energy as it drifts through the wire from b to a. . a& b& A current I = 2.5 A is flowing from left to right through a straight aluminum wire with a circular cross section of diameter D=(0.3 / π1/2) mm and length L=5.0 m, as shown above (not drawn to scale!). Aluminum has a resistivity of 2.7×10-8 Ω m (at room temperature). (f) Suppose in some other galaxy, far, far away, the wire was made of anti-matter. The charge carriers in this “anti-aluminum” wire are then anti-electrons, a.k.a. positrons, each positron having a positive charge, q = +e > 0. Assuming the same direction and strength of the electric current, I, the same wire length and diameter, and the same resistivity as before, answer questions (a)-(e) for the anti-aluminum wire and, respectively, for the “conduction positrons” flowing through the wire. Note: You don’t have to re-do any of the numerical calculations. Instead just state for each part, (a)-(e), whether, or not, the answer for the anti-aluminum wire and its positron charge carriers is the same as for the aluminum wire with electron charge carriers; and, if not, how the answer changes. In particular, how do the answers change, or not, for parts (d) and (e)? Since the strength and direction of the current flow in the anti-‐aluminum wire, the resistivity and the wire size is the same as before, the wire resistance, R, and the potential difference Vab = +15V > 0 is the same as in part (a) as before. For the same reasons, the heating power, P, and the amount of heat generated during 5h of current flow, H5h = 675 kJ is the same as in part (b); and the electric field E = 3.0 V/m, with E pointing from left to right, is the same as in part (c). However, because the positrons have positive charge, q=+e>0, the electric force acting on the positrons F = q E = (+e) E is in the same direction as the E-‐vector. So, the F-‐vector acting on the positrons points from left to right. Hence the positrons‘ drift motion is from left to right, opposite to electrons’ drift motion part (d) Since the positron moves through the wire from left to right, i.e., from a to b, its change in potential energy is ΔU = Ub – Ua = q (Vb – Va) = q (−Vab )= (+1.6×10-‐19 C) (−15V) = −2.4×10-‐18 J < 0 , which is the same as for the electron, moving from right to left, in part (e), using q=+e=+1.6×10-‐19C for the positron charge. The fact that ΔU<0 means that the positron loses potential energy, again the same as for the electron in part (e), as the positron drifts through the wire from a to b. Notice: The positron charge, q=+e, is opposite in sign to the electron charge. However, the positron also drifts through the wire in the opposite direction as the electron. Hence, the sign of the electric potential difference traversed by the positron, Vb – Va = −Vab = −15V, is opposite in sign to the potential difference traversed by the electron, Va – Vb = +Vab = +15V. Hence the potential energy change , ΔU, i.e., the product of carrier charge, q, and traversed electric potential difference, ΔV, is the same, in sign and magnitude, for electrons and positrons. a& b& A current I = 2.5 A is flowing from left to right through a straight aluminum wire with a circular cross section of diameter D=(0.3 / π1/2) mm and length L=5.0 m, as shown above (not drawn to scale!). Aluminum has a resistivity of 2.7×10-8 Ω m (at room temperature). . Returning to our own galaxy and planet earth… Exercise&1&(cont’d.)&& (g) Calculate the total number of electrons, N5h , flowing through the aluminum wire during 5h. [Hint review the definition of electric current, I, and recall how much charge, q, each electron carries.] I I Recall (h) that the urrent I is the rate at which charge Q flows though the wire, i.e., From Nc 5h found in part (g) and ΔU found in part (e), calculate the combined total I = or dQ/dt gain loss of electric potential energy, ΔUtot , from all electrons flowing through the during 5h.tCompare ΔUtot toothe result from part (b):passed the totaltheat energy generated where wire Q(t) is the otal amount f charge that has hrough the wire up to time in 5h by the current flow. t. So, for a constant, i.e., time-‐independent, current, I, the total charge, Q a& b&5h, passing through Atcurrent he wire d uring t =5h i s: I = 2.5 A is flowing from left to right through a straight aluminum wire with a Qcircular × t = (2.5A) (5h) (3600s/h) 45,000 C. L=5.0 m, as shown above 5h = Icross section of diameter D=(0.3 / π1/2)=mm and length -8 Ω m (at room temperature). (not drawn to scale!). Aluminum has a resistivity of 2.7×10 The carriers transporting this charge are electrons, each electron carrying an -‐19 absolute charge of |q|= e = 1.6×10 C. Hence, the total number of carriers that have Returning to our own galaxy and planet earth… passed through the wire during t=5h is therefore: +3C) / (N -‐19 through NCalculate / total q = number (45 ×of 10electrons, 1.6×10 = 2the .8125 ×1023 electrons (g) aluminum wire 5h = Q5hthe 5h , flowing C) during 5h. [Hint review the definition of electric current, I, and recall how much . charge, q, each electron carries.] (h) From N5h found in part (g) and ΔU found in part (e), calculate the combined total gain or loss of electric potential energy, ΔUtot , from all electrons flowing through the wire during 5h. Compare ΔUtot to the result from part (b): the total heat energy generated in 5h by the current flow. ΔUtot = N5h × ΔU = (2.8125 ×1023) (−2.4×10-‐18 J) = −675 kJ . As expected, the total potential energy lost by all electrons, −ΔUtot, passing through the wire during 5h, is exactly the same as the total amount of heat energy generated in the wire during those 5h. . a& b& A current I = 2.5 A is flowing from left to right through a straight aluminum wire with a circular cross section of diameter D=(0.3 / π1/2) mm and length L=5.0 m, as shown above (not drawn to scale!). Aluminum has a resistivity of 2.7×10-8 Ω m (at room temperature). (i) Calculate the conduction electron number density, ne , in aluminum, that is, the number of conduction electrons per volume.Exercise&1&(cont’d.)&& Use this: Since aluminum is trivalent (in case you already know any chemistry), each aluminum provides 3 conduction electrons. The molar mass of aluminum is 27 g per mole of Iatom aluminum atoms. The number of aluminum atoms in one mole is given by Avogadro’s number, NA =6.02×1023 atoms per mole. The mass density of aluminum is 2.70 g/cm3. I . (j) Calculate the electron drift speed, that is, the mean velocity of the electrons’ motion 3: in part (i), the current I, the wire cross section to the current [Hint:Vuse ne found Mass of due aluminum in flow. volume =1cm a& b& and=the charge, q = 3-e.] M ρmelectron’s V = (2.70g/cm ) (1cm3) = 2.70 g A current I = 2.5 A is flowing from left to right through a straight aluminum wire with a 3: Number circular of moles f aluminum in vD=(0.3 olume =1cm crossosection of diameter / πV1/2 ) mm and length L=5.0 m, as shown above -8 Ω N(not M / to Mscale!). 2.70g) /has (27 g/mole) 0.100 mmole m =drawn mol = (Aluminum a resistivity of =2.7×10 (at room temperature). Number of aluminum atoms in volume V=1cm3: (i) Calculate the conduction electron number density,23ne , in aluminum, that is, the number N = Nm×N mole)(6.02×10 atoms/mole)=6.02 × 1022atoms A= (0.100 ofatoms conduction electrons per volume. Use this: Number of conduction electrons in volume V=1cm3: 22 electrons N = trivalent 3 × (6.02 × 1you 022already ) = 18.06 10chemistry), Since aluminum (in case know×any each aluminum e=3×N atoms is atom 3 conduction electrons. molar mass of aluminum is 27 g per mole of Number of cprovides onduction electrons per vThe olume: aluminum atoms. The number of22aluminum atoms in one mole is22 given -‐3 by Avogadro’s 3 28 -‐3 n e = Ne/V = (18.02 × 10 )/(1cm ) = 18.06 × 10 cm = 18.06 × 10 m . number, NA =6.02×1023 atoms per mole. The mass density of aluminum is 2.70 g/cm3. . (j) Calculate the electron drift speed, that is, the mean velocity of the electrons’ motion due to the current flow. [Hint: use ne found in part (i), the current I, the wire cross section and the electron’s charge, q = -e.] . Current density in the wire: J = I / A = (2.5 A) / (2.25 × 10-‐8 m2) = 1.111 × 108 A/m2 using wire cross-‐section area, A=2.25 × 10-‐8 m2, from part (a). Current density, J, is given in terms of drift speed, vDrift, carrier number density, ne, and carrier charge, |q| , see FlipItPhysics or textbook, by: J = |q| ne vDrift vDrift = J / ( |q| ne ) = (1.111 × 108 A/m2) / [ (1.6 × 10-‐19 C) (18.06 × 1028 m-‐3 ) ] = 0.00384 m/s = 3.84 mm/s .