Electric Current in Conductors
Exercises, Q1 - Q10
1. The amount of charge passed in time t through a cross-section of a wire is
Q(t) =At²+Bt+C
(a) Write the dimensional formula for A, B, and C.
(b) If the numerical values of A, B and C are 5, 3 and 1 respectively in SI units, find the value of the current at t =5 s.
ANSWER: (a) The left-hand side is the charge. The dimensional formula of charge is = [IT].
To make this given formula dimensionally correct each term on the right-hand side must have this dimensional formula. Hence,
At² =[A][T²] should be equal to [IT],
→[A][T²] =[IT]
→[A] =[IT⁻¹]
Similarly Bt → [B][T]
i.e. [B][T] =[IT]
→[B] = [I].
And the third term, [C] =[IT].
(b) With the given value the formula becomes,
Q(t) = 5t² +3t +1,
The current through the cross-section,
i(t) =dQ/dt = 10t +3.
Hence current at t = 5 s,
i(5) =10*5 +3 =53 A.
2. An electron gun emits 2.0x10¹⁶ electrons per second. What electric current does this correspond to?
ANSWER: The charge on an electron is e =1.602x10⁻¹⁹ C.
The number of electrons emitted per second, n =2.0x10¹⁶
Hence the current = charge flowing per second
=2.0x10¹⁶*1.602x10⁻¹⁹
=3.2x10⁻³ A.
3. The electric current existing in a discharge tube is 2.0 µA. How much charge is transferred across a cross-section of the tube in 5 minutes?
ANSWER: Current, i =2.0 µA
→i =2.0x10⁻⁶ A.
Time, t = 5 minutes =5*60 s =300 s.
Hence the charge transferred in this time,
Q =i*t =2.0x10⁻⁶*300 C
=6.0x10⁻⁴ C.
4. Th current through a wire depends on time as
i = i₀ +αt,
where i₀ = 10 A and α =4 A/s. Find the charge crossed through a section of the wire in 10 seconds.
ANSWER: i = iₒ +αt
The charge crossed through a section in t seconds,
Q(t) =∫i dt
=∫(iₒ +αt) dt
=[iₒt +αt²/2 +C]
{between limits 0 to t, C is integration constant}
Putting the limits,
Q(t) =[iₒt+½αt²+C -C]
=iₒt +½αt²
=10t +½*4t²
=10t +2t²
Hence the charge crossed through the section in 10 s is
Q =10*10 +2*10²
=100 +200 =300 C.
5. A current of 1.0 A exists in a copper wire of a cross-section of 1.0 mm². Assuming one free electron per atom calculate the drift speed of the free electrons in the wire. The density of copper is 9000 kg/m³.
ANSWER: Given that, i = 1.0 A,
Area of cross-section, A =1.0 mm²
→A =1.0x10⁻⁶ m².
Charge of electron, e =1.602x10⁻¹⁹ C.
The drift speed of an electron in the conductor, vd =i/neA, where n = number of electrons per unit volume.
With the given condition number of atoms of copper is also n. The atomic weight of copper =63.55, hence 63.55 gm of copper will have 6.02x10²³ atoms.
The volume of this much copper =63.55x10⁻³/9000 m³
=7.06x10⁻⁶ m³
n = number of electrons per unit volume
=6.02x10²³/7.06x10⁻⁶
=8.52x10²⁸ /m³
Hence the drift speed,
vd =i/neA
=1.0/(8.52x10²⁸*1.602x10⁻¹⁹*10⁻⁶)m/s
=7.33x10⁻⁵ m/s
=7.33x10⁻² mm/s
=0.073 mm/s.
6. A wire of length 1 m and radius 0.1 mm has a resistance of 100 Ω. Find the resistivity of the material.
ANSWER: Length of the wire, l = 1 m.
Area of the cross-section of the wire,
A = πr² =π(1x10⁻⁴)² m²
=πx10⁻⁸ m²
R =100 Ω
Hence the resistivity of the material,
ρ =AR/l
=πx10⁻⁸*100/1 Ω-m
=πx10⁻⁶ Ω-m.
7. A uniform wire of resistance 100 Ω is melted and recast in a wire of length double that of the original. What would be the resistance of the wire?
ANSWER: Resistance of the wire, R =100 Ω.
If the resistivity of the wire =ρ
R =ρl/A
→ρ =RA/l.
Since the volume of wire in both cases is constant = Al.
When l' =2l
New area, A' =Al/l' =Al/2l =A/2
Hence the new resistance,
R' =ρl'/A'
=(RA/l)*2l/(A/2)
=4R
=4*100 Ω
=400 Ω.
8. Consider a wire of length 4 m and cross-sectional area 1 mm² carrying a current of 2 A. If each cubic meter of the material contains 10²⁹ free electrons, find the average time taken by an electron to cross the length of the wire.
ANSWER: Length, l =4 m.
Area of cross-section, A = 1 mm²
→A =1x10⁻⁶ m².
Current, i =2 A
Number of free electrons per unit volume, n =10²⁹
Hence the drift speed,
vd =i/neA
=2/(10²⁹*1.602x10⁻¹⁹*1x10⁻⁶)
=1.25x10⁻⁴ m/s
Hence the average time taken by an electron to cross the length of the wire =4/1.25x10⁻⁴ s
=3.2x10⁴ s.
=3.2x10⁴/3600 hours
≈8.9 hours.
9. What length of a copper wire of cross-sectional area 0.01 mm² will be needed to prepare a resistance of 1 kΩ? The resistivity of copper = 1.7x10⁻⁸ Ω-m.
ANSWER: Resistivity, ρ =1.7x10⁻⁸ Ω-m.
Cross-sectional area, A =0.01/10⁶ m²
→A =1x10⁻⁸ m².
Required resistance, R = 1 kΩ =1000 Ω
Hence the needed length of wire,
l =AR/ρ
=1x10⁻⁸*1000/(1.7x10⁻⁸) m
=588 m
=0.588 km ≈ 0.60 km.
10. Figure (32-E1) shows a conductor of length l having a circular cross-section. The radius of cross-section varies linearly from a to b. The resistivity of the material is ρ. Assuming that b-a <<l, find the resistance of the conductor. 
The figure for Q-10
ANSWER: Consider a section at distance x from the smaller end and a very small length dx. Let the radius at this section =r. We equate the slope of the outer length with the axis. 
The figure for Q-10
tanß =(r-a)/x
also tanß =(b-a)/l. Equating,
(r-a)/x =(b-a)/l
→(b-a)x =rl-al.
Differentiating both sides,
(b-a)dx =l.dr
dx ={l/(b-a)}dr
The resistance of this small length dx of the wire,
dR =ρ*dx/A
=ρ*{l.dr/(b-a)}/πr²
= {ρl/π(b-a)}*{dr/r²}
Hence, R =∫dR
={ρl/π(b-a)}∫(1/r²)dr
={ρl/π(b-a)}[(-1/r)+C]
Putting the limits between a and b,
R ={ρl/π(b-a)}*[(-1/b) -(-1/a)]
={ρl/π(b-a)}*[1/a -1/b]
={ρl/π(b-a)}*[(b-a)/ab]
=ρl/πab.
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Links to the Chapters
Links to the Chapters
CHAPTER- 32- Electric Current in ConductorsCHAPTER- 31- CapacitorsCHAPTER- 30- Gauss's Law
CHAPTER- 29- Electric Field and Potential
CHAPTER- 28- Heat Transfer
OBJECTIVE -I
CHAPTER- 26-Laws of Thermodynamics
CHAPTER- 25-CALORIMETRY
Questions for Short Answer
OBJECTIVE-I
OBJECTIVE-II
EXERCISES - Q-11 to Q-18
CHAPTER- 24-Kinetic Theory of Gases
CHAPTER- 23 - Heat and Temperature
CHAPTER- 21 - Speed of Light
CHAPTER- 20 - Dispersion and Spectra
CHAPTER- 19 - Optical Instruments
CHAPTER- 18 - Geometrical Optics
CHAPTER- 17 - Light Waves
CHAPTER- 16 - Sound Waves
CHAPTER- 15 - Wave Motion and Waves on a String
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
CHAPTER- 11 - Gravitation
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 32- Electric Current in Conductors
CHAPTER- 31- Capacitors
CHAPTER- 30- Gauss's Law
CHAPTER- 29- Electric Field and Potential
CHAPTER- 28- Heat Transfer
CHAPTER- 26-Laws of Thermodynamics
CHAPTER- 25-CALORIMETRY
Questions for Short Answer
OBJECTIVE-I
OBJECTIVE-II
CHAPTER- 24-Kinetic Theory of Gases
CHAPTER- 23 - Heat and Temperature
CHAPTER- 21 - Speed of Light
CHAPTER- 20 - Dispersion and Spectra
CHAPTER- 19 - Optical Instruments
CHAPTER- 18 - Geometrical Optics
CHAPTER- 17 - Light Waves
CHAPTER- 16 - Sound Waves
CHAPTER- 15 - Wave Motion and Waves on a String
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
CHAPTER- 11 - Gravitation
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 8 - Work and Energy
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CHAPTER- 7 - Circular Motion
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CHAPTER- 5 - Newton's Laws of Motion
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CHAPTER- 4 - The Forces
The Forces-
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CHAPTER- 3 - Kinematics - Rest and Motion
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CHAPTER- 2 - "Physics and Mathematics"
CHAPTER- 2 - "Physics and Mathematics"
