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Chapter 31, Problem 1P
To determine

The average induced emf in the circuit for the given time interval.

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Answer to Problem 1P

The average induced emf for the given time interval is 100V .

Explanation of Solution

Given Info: The inductance of the given inductor is 2.00H , steady current is 0.500A and time interval is 10.0ms .

Formula to calculate induced emf is,

ε=Ldidt (1)

Here,

ε is the induced emf in the circuit.

L is the inductance of inductor.

didt is the rate of change in the current.

Negative sign indicates the induced emf is opposite to the direction of current.

The expression for the rate of change of current is,

didt=i2i1t (2)

Substitute 0.500A for i2 , 0A for i1 , 10ms for t . in equation (2).

didt=0.500A0A10ms

didt=0.500A10ms(103s1ms)=0.500A10×103s=50A/s

Substitute 2.00H for L and 50A/s for didt in equation (1) to calculate induced emf.

ε=Ldidt=2.00H×50A/s=100V

Conclusion:

Therefore, the induced emf in the inductor during the given time interval is 100V .

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Chapter 31 Solutions

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

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