Download Conduction Electron Per Atom - Physics with Calculus - Solved Exam and more Exams Calculus in PDF only on Docsity!
Prof. Yasu Takano Prof. Paul Avery Oct. 17, 2007
Exam 2 Solutions
- ( WebAssign 26.6 ) A current of 1.5 A flows in a copper wire with radius 1.25 mm. If the cur- rent is uniform, what is the electron drift speed in m/s? Copper has a density of 8960 kg/m^3 , atomic mass of 63.54 and one conduction electron per atom.
(1) 2.2 × 10 −^5
(2) 3.6 × 10 −^4
(3) 5.4 × 10 −^3
(4) 7.4 × 10 −^4
(5) 3.0 × 10 −^5
The current density j is given by j = i / π r^2 = en ve d, where ne is the conduction electron density
and v (^) d is the drift velocity. The electron number density for copper is the same as the atomic den- sity since there is one conduction electron per atom. That calculation gives the electron number
density ne = ρ Cu N 0 /10 −^3 A = 8960 × ( 6.02 × 10 23 )/ 0.06354 = 8.49 × 10 28 /m^3 , where the factor of
10^ −^3 is the conversion from g to kg in the atomic mass. Using j = i / π r^2 , we obtain vd = 2.2 ×
10^ −^5 m/s.
- ( WebAssign 26.41 ) A heating element is made by maintaining a potential difference of 60.0 V along the length of a Nichrome wire with radius 0.945 mm and a resistivity of 5.00 × 10 −^7 Ω m. If the element dissipates 5000 W, what is its length in m?
(1) 4. (2) 5. (3) 2. (4) 0. (5) 0.
The power dissipated is P = i R^2 = V^2 / R, where R = ρ L / A, ρ is the resistivity, L is the length
and A = π r 2 is the cross sectional area. Putting in all the numbers yields L = 4.0 m.
- ( WebAssign 27.47 ) A 18.0 Ω resistor and a capacitor are connected in series and then a 13.0 V potential difference is suddenly applied across them. The potential difference across the capaci- tor rises to 7.00 V in 1.50 μs. What is the capacitance in μF?
(1) 0. (2) 2. (3) 4. (4) 6. (5) 1.
The voltage across an initially uncharged capacitor is V ( ) t = V 0 ( 1 − e − t^ / RC ) , where V 0 = 13 is
the applied voltage. Rearranging yields the equation exp ( −1.5 × 10 −^6 / RC )= 1 − V / V 0 = 0..
Taking the log of both sides and solving for C yields C = 0.11 μ F.
- ( WebAssign 27.33 ) Consider the circuit shown in the accompanying figure. What is the power dissipated (in watts) in the 5 Ω resistor?
(1) 0. (2) 0. (3) 0. (4) 2. (5) 0.
The two equations from Kirchoff’s rules are 1 − 2 i 1 − 5 ( i 1 + i 2 )= 0 and 2 − i 2 − 5 ( i 1 + i 2 )= 0.
Solving yields i 1 (^) + i 2 = 5 /17 = 0.294 A. So the power is 0.294^2 × 5 = 0.43 W.
i 1 i 2
- ( Lecture problem ) A uniform current flows in a long straight copper wire whose radius is 5. mm. Inside the wire, at a distance of 1.0 mm from its center, the magnetic field due to the current is 3.0 × 10 −^4 T. What is the current in A?
(1) 37. (2) 260 (3) 105 (4) 65. (5) 18.
From Ampere’s law, B ( 2 π r ) = μ0 enc i , where i enc is the enclosed current. Solving for ienc yields
i (^) enc = 1.5 A. Since the current is uniform throughout the wire cross section, the total current i can be found from the ratio of areas or i = 5^2 × i (^) enc = 37.5 A.
- ( WebAssign 29-29 ) Four long parallel wires are arranged on a plane, as shown in the attached figure, with 2.0 cm gaps between them. Each wire carries a 3.0 A current in the direction indi- cated by the arrow. On the wire labeled C, what is the magnetic force per meter in N/m?
(1) 4.5 × 10 −^5 (2) 1.4 × 10 −^4 (3) 3.0 × 10 −^3 (4) 2.3 × 10 −^4 (5) 6.5 × 10 −^2
We first note that wire C is repelled from both B and D, so there is no net effect from them. The only net force is with A. A quick calculation yields the value of the force per unit length as
( ) (^ )
7 5
F / L μ 0 i iA C / 2 π d 4 π 10 3 3 / 2 π 0.04 4.5 10
= = × −^ × × × = × − N/m.
- ( Chapter problem 29.53 ) A wire carrying a 4.0 A current forms a circular loop, 5.0 mm in radius. At the center of the loop is a very tiny current loop, radius 98 nm and current 1 μA, ori- ented perpendicular to the direction of the magnetic field produced by the larger loop. What is the torque on the smaller loop in N m?
(1) 1.5 × 10 −^23
(2) 1.2 × 10 −^19
(3) 7.5 × 10 −^20
(4) 6.1 × 10 −^21
The torque τ is τ = μ B, where μ is the dipole moment of the inner current and B is the B field
generated by the outer current near the center. Here μ= A i in in (^) = π r i in in^2 and B = μ0 out i (^) / 2 r out ,
where “in” and “out” refer to the respective loops. Solving yields τ = 1.5 × 10 −^23 N m.
