HW05 solutions - General Physics: Modern Physics PDF

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General Physics: Modern Physics
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Homework(5(solutions( ! Notes:!Please!keep!in!mind!that!most!of!the!problems!in!MasteringPhysics!with!numerical!answers!are! randomized!for!each!student.!The!solutions!below!correspond!to!the!version!of!the!problem!that!is!in!the! paper!copy!of!the!textbook.!There!may!be!some!differences!between!the!textbook!version!and!the!online! version.! 1. Reading!questions! a. This!is!not!true.!Bring!a!compass!near!the!current.!The!compass!needle!does!not!point!towards!the! current!or!away!from!it.!If!you!use!the!compass!to!follow!the!magnetic!field!lines!produced!by!the! current!you!find!they!form!closed!paths!around!the!current.!The!right!hand!rule!for!the!magnetic! field!determines!the!direction!of!these!field!lines!(clockwise!or!counterclockwise).! b. This!is!true.!If!the!external!magnetic!flux!through!a!wire!coil!is!increasing,!the!induced!current!will! produce!a!magnetic!field!that!whose!flux!is!in!the!opposite!direction.!If!the!external!magnetic!flux! through!a!wire!coil!is!decreasing,!the!induced!current!will!produce!a!magnetic!field!that!whose! flux!is!in!the!same!direction.!See!Figure!18.8!on!pg.!669!of!the!textbook!for!examples.! 2. Problem 17.27 The magnetic field under the power line is

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)

4π × 10−7 T ⋅ m/A (500 A ) µ0 I = B= = 1.0 × 1025 T 2π (10 m ) 2πr The direction is north. 3. Problem 17.37 Sketch and translate The system with two parallel wires carrying currents in opposite direction is shown to the right. Represent mathematically The magnetic field at a perpendicular distance r from a long, straight current-carrying wire is given by B = µ 0 I /2πr, where µ 0 = 4π × 10−7 T ⋅ m/A.  With two wires, the net B field is the vector sum of the individual field contributions. Solve and evaluate (a) The net magnetic field at the mid-point A is the sum of the fields from each current: 4µ I µ 0 I1 µ 0 I1 + = 0 1 BA = B1 + B2 = 2π ( d /2 ) 2π ( d /2 ) 2πd =

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4 4π × 10−7 T ⋅ m/A (30 A ) 2π ( 2.0 m )

The direction is out of the page. (b) Next, at point P shown in the figure, the magnetic field is

= 1.2 × 1025 T

BP = B1 + B2 = 2 =2

µ 0 I1 2π ( d /2 )

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+

µ 0 I1 = 2 2µ 0 I1 3πd 2π (3d /2 )

)

2 4π × 10−7 T ⋅ m/A (30 A ) 3π ( 2.0 m )

= 24.0 × 1026 T

The negative sign means that the field points into the page. 4. Tutorial (the problem parts plus the hints in MasteringPhysics walk you through how to do it) 5. Problem 18.61 (a) The induced emf is εin =|

! 1.5 T $ ΔΦ ΔB |=| | A = # & 0.30 m × 0.40 m = 0.36 V. Δt Δt " 0.50 s %

(

(b) With R = 5.0 Ω, the induced current is I in =

)

ε in 0.36 V = = 0.072 A. 5.0 Ω R

(c) The direction of the induced current is clockwise, as viewed from above. 6. Problem 18.22 To achieve an emf of 1.5 V, the magnitude of the rate of change of magnetic field must be Δ𝐵 Δ 𝐵𝐴 1 Δ 𝐵𝐴𝑐𝑜𝑠𝜃 ΔΦ( = 𝑁𝐴 =𝑁 =𝑁 𝜀"# = 𝑁 Δ𝑡 Δ𝑡 Δ𝑡 Δ𝑡 ΔB εin 1.5 V = = = 50 T/s Δt NA 100 3.0 × 10−4 m 2

( )(

)

Since ΔB = 0.040 T, the time interval is Δt = 8.0 × 1024 s. That is, in order to make the bulb light, you would need to move the magnet 4.0 cm in Δt = 8.0 × 1024 s, which implies a speed of 50 m/s, or about 110 mph. This speed is unreasonable. 7. Problem 18.23 Represent mathematically Faraday’s law of induction states that the average induced emf ε in in a coil with N loops is the magnitude of the rate of change of the magnetic flux | ΔΦ/Δt | multiplied by N: εin = N | this problem, the area is changing with time. Solve and evaluate The average induced emf is

εin = N | !

ΔA ΔΦ |= NB | | cos θ = 300 5.0 × 10−5 T Δt Δt

( )(

# 42 × 10−4 m 2 & % (cos53° = 1.9 × 10−5 V 2.0 s $ '

)

ΔΦ | . In Δt...