The current i through a 4.6 H inductor varies with time t as shown by the graph of Fig. 32-27. T

The current i through a 4.6 H inductor varies with time t as...

The current $i$ through a $4.6 \mathrm{H}$ inductor varies with time $t$ as shown by the graph of Fig. $32-27$. The inductor has a resistance of $12 \Omega$. Find the magnitude of the induced emf $\mathscr{C}$ during the time intervals (a) $t_{1}=0$ to $t_{2}=2 \mathrm{~ms}$, (b) $t_{2}=2 \mathrm{~ms}$ to $t_{3}=5 \mathrm{~ms}$, (c) $t_{3}=5 \mathrm{~ms}$ to $t_{4}=6 \mathrm{~ms}$. (Ignore the behavior at the ends of the intervals.)

Key Concepts

Self-Inductance

Self-inductance is a property of an inductor that quantifies its ability to store magnetic energy when current flows through it. It is defined by the ratio of the magnetic flux to the current that produces it, and it directly determines how strongly the inductor opposes changes in current. A larger inductance means that a given rate of change in current will produce a larger induced electromotive force (emf).

Faraday's Law of Induction

Faraday's Law of Induction states that a changing magnetic flux through a circuit induces an electromotive force (emf) in the circuit. In the context of an inductor, this means that a time-varying current, which produces a changing magnetic field, gives rise to an induced emf that is proportional to the rate of change of current, mathematically expressed as emf = -L (di/dt).

Rate of Change of Current

The rate of change of current (di/dt) is a measure of how quickly the current in a circuit changes over time. In problems where current varies over specific time intervals, the slope of the current versus time graph (?i/?t) is used to compute the induced emf in the inductor. The magnitude of the induced emf is determined by multiplying this rate by the inductance, while the sign indicates that the induced emf opposes the change in current.

Transcript

00:01 Hi in the given problem there is an inductor coil whose self inductance is given as l is equal to 4 .6 henry then there is a curve representing the variation of current passing through this inductor coil the passage of time here this time is measured in milliseconds and this current has been measured in amperors and roughly if we study this curve we conclude that initially the current increases from 0 to 7 ampere here this is 7 for a time interval of from 0 to 2 second then this current drops from 7 ampere to 5 amp over a time interval from 2 milliseconds to 5 milliseconds and then finally it drops to 0 in another 1 milliseconds so this is 6 milliseconds here now using the expression for emf induced across inductor that is given as along with the sign convention this is negative of l times of the i by d t the time rate of change of current and here with the help of this curve we can say d i by d t is nothing but the slope of i versus the curve and we know slope is given as perpendicular upon base so everywhere so in the first time interval here perpendicular upon base will be of this triangle then in the second time interval this triangle will be used and in the third time interval this triangle will be used so in the first part of the problem we have to find emf induced in a time interval from e1 equals to zero to 2 equals 2 milliseconds means the first triangle here will be used hence mf induced minus for l this is 4 .6 henry for slope perpendicular upon base for perpendicular this is 7 minus 0 ampere divided by base which is 2 milliseconds or 2 into 10 each per minus 3 second minus 0 second and here it will be given by minus 1 .61 into 10 dash 4 fold which is answer for the first part of the problem now in the second part of the problem time interval is from t2 equals to 2 milliseconds to 3 which is equal to 5 milliseconds again using the same expression minus 4 .6 for self -inductance now for perpendicular this is 5 minus 7 ampere and divided by 5 milliseconds means 5 into 10 dash per minus 3 minus 2 2 2 into 10 x 4 minus 2 second so finally this cmf induced here comes out to be 3 .1 it will become positive 3 .1 into 10 dash 3 volt another answer for the given problem and in the third part of the problem the time limit from t3 equals to 5 milliseconds to t4 is equal to 6 milliseconds so finally here it will be given by minus 4 .6 again and the final time is final current is 0 and initially it was 5.

04:37 So the height of perpendicular 0 minus 5 divided by time, which is 6 milliseconds or 6 into 10 minus 3 minus 0.

04:47 Sorry, this will be here.

04:49 If you look this, this is 5, 6 minus 5 for the base 5 into 10 inch per minus 3...

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