Cap 30 fisisca 3

1. The amplitude of the induced emf in the loop is

ε m = Aµ 0 ni0ω = (6.8 ×10−6 m 2 )(4π × 10 −7 T ⋅ m A)(85400 / m)(1.28 A)(212 rad/s)
= 1.98 ×10−4 V.

2. (a) ε =

dΦ B d = 6.0t 2 + 7.0t = 12t + 7.0 = 12 2.0 + 7.0 = 31 mV. dt dt

c

h

b g

(b) Appealing to Lenz’s law (especially Fig. 30-5(a)) we see that the current flow in the loop is clockwise. Thus, the current is to left through R.

3. (a) We use ε = –dΦB/dt = –πr2dB/dt. For 0 < t < 2.0 s:

ε = −πr 2

dB 2 0.5T = −π ( 0.12m ) = −1.1×10−2 V. dt 2.0s

(b) 2.0 s < t < 4.0 s: ε ∝ dB/dt = 0. (c) 4.0 s < t < 6.0 s:

ε = − πr 2

dB = − π 012 m . dt

b

. g FGH 6.0−s0−.54T.0sIJK = 11 × 10
2

−2

V.

4. The resistance of the loop is
R=ρ

π ( 0.10 m ) L = (1.69 × 10−8 Ω ⋅ m ) = 1.1× 10−3 Ω. −3 2 A π ( 2.5 × 10 ) / 4

We use i = |ε|/R = |dΦB/dt|/R = (πr2/R)|dB/dt|. Thus
−3 dB iR (10 A ) (1.1×10 Ω ) = = = 1.4 T s. 2 dt π r 2 π ( 0.05 m )

5. The total induced emf is given by

ε = −N

dΦB dB d di di = − NA = − NA ( µ 0 ni ) = − N µ 0 nA = − N µ 0 n(π r 2 ) dt dt dt dt dt
2

= −(120)(4π ×10 −7 T ⋅ m A)(22000/m) π ( 0.016m ) = 0.16V.

1.5 A 0.025 s

Ohm’s law then yields i =| ε | / R = 0.016 V / 5.3Ω = 0.030 A .

6. Using Faraday’s law, the induced emf is d ( πr 2 ) d ( BA ) dΦB dA dr =− = −B = −B = −2πrB ε =− dt dt dt dt dt = −2π ( 0.12m )( 0.800T )( −0.750m/s ) = 0.452V.

7. The flux Φ B = BA cosθ does not change as the loop is rotated. Faraday’s law only leads to a nonzero induced emf when the flux is changing, so the result in this instance is 0.

8. The field (due to the current in the straight wire) is out-of-the-page in the upper half of the circle and is into the page in the lower half of the circle, producing zero net flux, at any time. There is no induced current in the circle.

9. (a) Let L be the length of a side of the square circuit. Then the magnetic flux through the circuit is Φ B = L2 B / 2 , and the induced emf is

εi = −

dΦB L2