
When the current in a coil changes from $8\text{ ampere}$ to $2\text{ ampere}$ in $3\times {{10}^{-2}}\text{ second}$, the e.m.f induced in the coil is $2\text{ volt}$. The self inductance of the coil in millihenry is
$A)\text{ }1$
$B)\text{ 5}$
$C)\text{ 20}$
$D)\text{ }10$
Answer
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Hint: This problem can be solved by using the direct formula for the e.m.f induced in the coil in terms of the change in current, the time in which the current changes and the self inductance of the coil. Using the formula and putting the values given in the question, we will get the self inductance of the coil.
Formula used:
$\varepsilon =-L\dfrac{\Delta I}{\Delta t}$
Complete answer:
We will use the formula for the emf induced in a coil due to the current changing in it.
The emf $\varepsilon $ induced in a coil of self inductance $L$ when there is a change of current $\Delta I$ flowing through it in a time period $\Delta t$ is given by
$\varepsilon =-L\dfrac{\Delta I}{\Delta t}$ --(1)
The negative sign implies that the emf induced tries to oppose the change in current in the coil.
Now, let us analyze the question.
Since, the current in the coil changes from $8A$ to $2A$, the change in the current in the coil is $\Delta I=2-8=-6A$.
The time period in which the current changes is $\Delta t=3\times {{10}^{-2}}s$
The emf induced in the coil due to this is $2V$.
Let the self inductance of the coil be $L$.
Therefore, using (1), we get
$2=-L\dfrac{-6}{3\times {{10}^{-2}}}=L\left( 2\times {{10}^{2}} \right)$
$\therefore L=\dfrac{2}{2\times {{10}^{2}}}=1\times {{10}^{-2}}=10\times {{10}^{-3}}H=10mH$ $\left( \because {{10}^{-3}}H=1mH \right)$
Therefore, we have found out the self inductance of the coil as $10mH$.
Therefore, the correct answer is $D)\text{ }10$.
Note:
Students must note that the self inductance of a coil is its characteristic property that determines how much opposition is offered by the coil to the change in current passing through it. A coil with a higher value of self inductance offers a greater opposition to the change in current through it and thus, produces a larger induced emf so that this large emf induced can oppose the change in the current in the coil.
Formula used:
$\varepsilon =-L\dfrac{\Delta I}{\Delta t}$
Complete answer:
We will use the formula for the emf induced in a coil due to the current changing in it.
The emf $\varepsilon $ induced in a coil of self inductance $L$ when there is a change of current $\Delta I$ flowing through it in a time period $\Delta t$ is given by
$\varepsilon =-L\dfrac{\Delta I}{\Delta t}$ --(1)
The negative sign implies that the emf induced tries to oppose the change in current in the coil.
Now, let us analyze the question.
Since, the current in the coil changes from $8A$ to $2A$, the change in the current in the coil is $\Delta I=2-8=-6A$.
The time period in which the current changes is $\Delta t=3\times {{10}^{-2}}s$
The emf induced in the coil due to this is $2V$.
Let the self inductance of the coil be $L$.
Therefore, using (1), we get
$2=-L\dfrac{-6}{3\times {{10}^{-2}}}=L\left( 2\times {{10}^{2}} \right)$
$\therefore L=\dfrac{2}{2\times {{10}^{2}}}=1\times {{10}^{-2}}=10\times {{10}^{-3}}H=10mH$ $\left( \because {{10}^{-3}}H=1mH \right)$
Therefore, we have found out the self inductance of the coil as $10mH$.
Therefore, the correct answer is $D)\text{ }10$.
Note:
Students must note that the self inductance of a coil is its characteristic property that determines how much opposition is offered by the coil to the change in current passing through it. A coil with a higher value of self inductance offers a greater opposition to the change in current through it and thus, produces a larger induced emf so that this large emf induced can oppose the change in the current in the coil.
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