17 worked answers · short steps first, explanations when you need them
Chapter 6 Exercises
Textbook pages 174–177
Question 1
Predict the induced-current direction in each of the six situations in Fig. 6.18.
Steps only
Use Lenz’s law, then the right-hand rule: (a) q → r → p → q; (b) p → r → q → p; (c) y → z → x → y; (d) z → y → x → z; (e) x → r → y → x; (f) no current.
Answer
(a) q → r → p → q. (b) p → r → q → p. (c) y → z → x → y. (d) z → y → x → z. (e) x → r → y → x. (f) No induced current.
Question 2
Use Lenz's law to find the induced-current direction when (a) an irregular wire loop becomes circular and (b) a circular loop is deformed into a narrow straight wire, as shown in Fig. 6.19.
Steps only
Use Lenz's law: oppose the flux increase in (a) and oppose the flux decrease in (b).
Answer
(a) Anticlockwise; (b) anticlockwise
Question 3
A solenoid has 15 turns per centimetre. A loop of area 2.0 cm² is placed inside it, normal to its axis. Its current rises from 2.0 A to 4.0 A in 0.1 s. Find the induced emf in the loop.
Steps only
ε=Aμ₀n(ΔI/Δt)=(2×10⁻⁴)(4π×10⁻⁷)(1500)(20)=7.54×10⁻⁶ V
Answer
7.54 μV
Question 4
A rectangular loop has sides 8 cm and 2 cm and moves out of a 0.3 T field, normal to its plane, at 1 cm/s. Find the emf and how long it lasts when the motion is (a) normal to the longer side and (b) normal to the shorter side.
A 1.0 m rod rotates at 400 rad/s about an axis through one end and perpendicular to the rod. A uniform 0.5 T field is parallel to the axis. Find the emf between the rod's ends.
Steps only
ε=½BωL²=½(0.5)(400)(1.0)²=100 V
Answer
100 V
Question 6
A 20-turn coil of radius 8.0 cm rotates at 50 rad/s in a 3.0×10⁻² T horizontal field. Find maximum and average induced emf, peak current for 10 Ω resistance, and average heating power. State the energy source.
Steps only
ε₀=NBAω≈0.603 V; I₀=ε₀/10≈0.0603 A; Pav=ε₀²/(2R)≈0.0182 W
Answer
0.60 V max; 0.060 A peak; 0.018 W average
Question 7
A 10 m east–west wire falls at 5.0 m/s, perpendicular to the horizontal component of Earth's field, 0.30×10⁻⁴ T. Find (a) the emf, (b) its direction and (c) the higher-potential end.
Steps only
ε=BLv=(0.30×10⁻⁴)(10)(5)=1.5 mV; v×B points east.
Answer
1.5 mV; east end positive
Question 8
Current falls from 5.0 A to zero in 0.1 s and the average induced emf is 200 V. Estimate the circuit's self-inductance.
Steps only
L=ε/(|ΔI|/Δt)=200/50=4.0 H
Answer
4.0 H
Question 9
Two adjacent coils have mutual inductance 1.5 H. Current in one changes from 0 to 20 A. Find the change in flux linkage of the other coil.
Steps only
Δ(NΦ)=MΔI=1.5×20=30 Wb-turn
Answer
30 Wb-turn
Question 10
A jet travels west at 1800 km/h. Its 25 m wing span is in a 5×10⁻⁴ T Earth's field with dip angle 30°. Find the voltage between the wing ends.
Steps only
ε=ℓvB sin30°=25(500)(5×10⁻⁴)(0.5)=3.125 V
Answer
3.1 V
Question 11
The loop in Exercise 6.4 is stationary while its 0.3 T field falls at 0.02 T/s. The loop is closed and has resistance 1.6 Ω. Find its heat power and source.
Steps only
ε=A|dB/dt|=3.2×10⁻⁵ V; P=ε²/R=6.4×10⁻¹⁰ W
Answer
6.4×10⁻¹⁰ W
Question 12
A 12 cm square loop moves at 8 cm/s in +x through a field in +z. The field increases by 10⁻³ T/cm toward −x and falls in time at 10⁻³ T/s. If loop resistance is 4.50 mΩ, find induced-current magnitude and direction.
Steps only
dB/dt along loop=−0.009 T/s; ε=A|dB/dt|=1.296×10⁻⁴ V; I=ε/R=28.8 mA, CCW viewed from +z
Answer
28.8 mA, counter-clockwise viewed from +z
Question 13
A 25-turn search coil of area 2 cm² is removed from a magnetic field (or turned 90°). It passes 7.5 mC through a total resistance of 0.50 Ω. Estimate the field strength.
Steps only
B=QR/(NA)=(7.5×10⁻³)(0.50)/[25(2×10⁻⁴)]=0.75 T
Answer
0.75 T
Question 14
A 15 cm rod slides at 12 cm/s on rails in a 0.50 T field; the closed-loop resistance is 9.0 mΩ. Find (a) emf and polarity, (b) charge build-up, (c) why electron forces balance when open, (d) retarding force when closed, (e) external power open/closed, (f) heat power and source, and (g) emf if the field is parallel to the rails.
Steps only
ε=Blv=9 mV; I=ε/R=1 A; F=BIl=0.075 N; P=Fv=I²R=9 mW; parallel B and v gives ε=0
Answer
9 mV; P positive; 0.075 N; 9 mW; zero if B∥rails
Question 15
An air-core solenoid is 30 cm long, has area 25 cm² and 500 turns. A 2.5 A current is switched off in 10⁻³ s. Find the average back emf.
Steps only
L=μ₀N²A/ℓ≈2.62 mH; ε=LΔI/Δt≈6.5 V
Answer
6.5 V
Question 16
(a) Find the mutual inductance between a long straight wire and a nearby square loop of side a, whose nearest side is distance x away (Fig. 6.21). (b) For I=50 A, a=0.10 m, x=0.20 m and loop speed 10 m/s away from the wire, find the induced emf.
Steps only
M=(μ₀a/2π)ln[(x+a)/x]; |ε|=(μ₀Iav/2π)[1/x−1/(x+a)]=1.67×10⁻⁵ V
Answer
M formula above; 17 μV
Question 17
A wheel of mass M and radius R has charge per unit length λ around its rim. A field B=−B₀k fills a circle of radius a<R inside the rim and is suddenly switched off. Find the wheel's angular velocity.