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A thin wire ℓ = 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
(a)
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Answer to Problem 31.78AP
Explanation of Solution
Given info: Length of wire is
The speed of the wire according to Newton’s law of motion can be given as,
Here,
Substitute
The distance covered by the wire can be given as,
Here,
Substitute
The total distance covered by wire can be given as,
Here,
Substitute
The magnetic field at a distance
Here,
The emf induced in the wire can be given as,
Here,
Substitute
Substitute
Conclusion:
Therefore, the equation for emf induced in the wire as function of time is
(b)
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Answer to Problem 31.78AP
Explanation of Solution
Given info: Length of wire is
The expression for emf can be given as in equation (1),
At
Conclusion:
Therefore, the minimum value of emf is
(c)
![Check Mark](/static/check-mark.png)
Answer to Problem 31.78AP
Explanation of Solution
Given info: Length of wire is
The expression for emf can be given as in equation (1),
From the above equation, at
Conclusion:
Therefore, the minimum value of emf is
(d)
![Check Mark](/static/check-mark.png)
Answer to Problem 31.78AP
Explanation of Solution
Given info: Length of wire is
The expression for emf can be given as in equation (1),
Substitute
Conclusion:
Therefore, the induced emf
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Chapter 31 Solutions
Physics for Scientists and Engineers, Technology Update (No access codes included)
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