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Chapter 30, Problem 47AP

A thin wire = 30.0 cm long is held parallel to and d = 80.0 cm above a long, thin wire carrying I = 200 A and fixed in position (Fig. P30.47). The 30.0-cm wire is released at the instant t = 0 and falls, remaining parallel to the current-carrying wire as it falls. Assume the falling wire accelerates at 9.80 m/s2. (a) Derive an equation for the emf induced in it as a function of time. (b) What is the minimum value of the emf? (c) What is the maximum value? (d) What is the induced emf 0.300 s after the wire is released?

Figure P30.47

Chapter 30, Problem 47AP, A thin wire  = 30.0 cm long is held parallel to and d = 80.0 cm above a long, thin wire carrying I =

(a)

Expert Solution
Check Mark
To determine

The equation for the emf induced in the wire as a function of time.

Answer to Problem 47AP

The equation for emf induced in the wire as function of time is (1.18×104)t0.8004.90t2

Explanation of Solution

Given info: Length of wire is 30.0cm , distance from parallel long wire is 80.0cm and current in the wire is 200A .

The speed of the wire according to Newton’s law of motion can be given as,

v=u+at

Here,

u is the initial speed of wire.

a is the acceleration of wire.

t is the time.

Substitute 0 for u and 9.80m/s2 for a in the above equation,

v=9.80t

The distance covered by the wire can be given as,

s=ut12gt2

Here,

s is the distance covered by the wire.

g is the acceleration due to gravity.

Substitute 0 for u and 9.80m/s2 for g in the above equation,

s=12(9.80m/s2)=4.90t2

The total distance covered by wire can be given as,

y=d+s

Here,

y is the total distance covered by the wire.

d is the distance between wires.

Substitute 80.0cm for d and 4.90t2 for s in the above equation,

y=(80.0cm)4.90t2=0.8004.90t2

The magnetic field at a distance y from the on a section of coil can be given as,

B=μ0I2πy

Here,

B is the magnetic field induced in the loop.

μ0 is the permeability constant.

I is the current induced in the loop.

y is any arbitrary distance from the wire.

The emf induced in the wire can be given as,

ε=Blv

Here,

ε is the emf induced in the wire.

l is the length of the wire.

v is the speed of the wire.

Substitute μ0I2πy for B in the above equation,

ε=(μ0I2πy)lv

Substitute 4π×107Tm/A for μ0 , 200A for I , 30.0cm for l , 9.80t for v and 0.8004.90t2 for y in the above equation,

ε=(4π×107Tm/A)(200A)(30.0cm)(1m100cm)(9.80t)2π(0.8004.90t2)=(1.18×104)t0.8004.90t2 (1)

Conclusion:

Therefore, the equation for emf induced in the wire as function of time is (1.18×104)t0.8004.90t2

(b)

Expert Solution
Check Mark
To determine

The minimum value of the emf.

Answer to Problem 47AP

The minimum value of emf is 0.

Explanation of Solution

Given info: Length of wire is 30.0cm , distance from parallel long wire is 80.0cm and current in the wire is 200A .

The expression for emf can be given as in equation (1),

ε=(1.18×104)t0.8004.90t2

At t=0 ,

ε=(1.18×104)×00.8004.90(02)=0

Conclusion:

Therefore, the minimum value of emf is 0.

(c)

Expert Solution
Check Mark
To determine

The maximum value of emf.

Answer to Problem 47AP

The maximum value of emf is infinity.

Explanation of Solution

Given info: Length of wire is 30.0cm , distance from parallel long wire is 80.0cm and current in the wire is 200A .

The expression for emf can be given as in equation (1),

ε=(1.18×104)t0.8004.90t2

From the above equation, at t= , ε= .

Conclusion:

Therefore, the minimum value of emf is

(d)

Expert Solution
Check Mark
To determine

The induced emf 0.300s after the wire is released.

Answer to Problem 47AP

The induced emf 0.300s after the wire is released is 98.3μV .

Explanation of Solution

Given info: Length of wire is 30.0cm , distance from parallel long wire is 80.0cm and current in the wire is 200A , instant of time is 0.300s .

The expression for emf can be given as in equation (1),

ε=(1.18×104)t0.8004.90t2

Substitute 0.300s for t in the above equation,

ε=(1.18×104)(0.300s)0.8004.90(0.300s)2=9.83×105V=98.3μV

Conclusion:

Therefore, the induced emf 0.300s after the wire is released is 98.3μV .

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Chapter 30 Solutions

Bundle: Physics For Scientists And Engineers With Modern Physics, 10th + Webassign Printed Access Card For Serway/jewett's Physics For Scientists And Engineers, 10th, Multi-term

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