Question

The current in the coil decreases from $10{\text{ }}Amp$ to ${\text{5 }}Amp$ in $0.1{\text{ }}\sec ond$. Calculate the coefficient of self-inductance if induced e.m.f. is $5{\text{ }}V$.

Answer 1

Hint:In the given question the values of current in the coil is given. We have to find out the difference of current in the coil. Then, by using the relation between induced emf and self-inductance we will find out the value of coefficient of self-inductance.

Formula used: $e = - L\dfrac{{dI}}{{dt}} - - - - \left( 1 \right)$ (negative sign gives the information of direction)
where $e = $ induced emf, $L = $ coefficient of self-inductance, $dI = $ change in current and $dt = $ change in time.

Complete step-by-step solution:
Induced emf (electromotive force) is defined as the generation of potential difference when there is a change in magnetic flux through the coil.
When there is a change in magnetic flux in a coil, there develops an opposite electromotive force known as self-inductance.
The relation between self-induction and induced emf is,
$e = - L\dfrac{{dI}}{{dt}} - - - - \left( 1 \right)$ (negative sign gives the information of direction)
where $e = $ induced emf, $L = $ co-efficient of self-inductance, $dI = $ change in current and $dt = $ change in time.
According to the question,
$dI = 5 - 10 = - 5$
$dt = 0.1{\text{ }}s$ and
$e = 5{\text{ }}V$
Substituting the above values in equation $\left( 1 \right)$ we get,
$5 = - L\dfrac{{ - 5}}{{0.1}}$
Arranging the equation we get,
$L = 0.1$
So, the co-efficient of self-inductance is $0.1{\text{ }}H$.

Additional information: Self inductance can also be expressed as $\phi = LI$ where $\phi = $ magnetic flux, $L = $ co-efficient of self-induction and $I = $current.


Note:It must be noted that change in current must be the value of current that became the last subtracted from the earlier first value. The unit of self-inductance is Henry. The negative sign in the formula suggests that the force of self-inductance is opposite to the induced emf.