Question

When electric current in a coil steadily changes from $+ 2{\text{A to - 2A}}$ in$0.05s$, an induced emf of $8.0V$ is generated in it. Then the self-induction of the coil is ________H

Answer 1

We have to find the self-induction of the coil. The emf of the coil, the change in current and time are provided in the question. It is a direct question. Use the self-induction formula directly to find the answer.

Complete step by step solution:
A battery and a switch are connected in series with a coil. When we turn on the switch, the current through the coil increases to maximum and the magnetic flux linked with the coil also gets increased. According to Lenz’s law an induced current flows through the coil opposes the further growth of current in the coil. When we turn off the switch the current through the coil decreases to a zero and the magnetic flux linked with the coil also decreases. The induced current will oppose the decay of current in the coil.
When the current in the coil changes, it enables it to produce an opposing induced emf in it. This property of a coil which produces an opposing induced emf is called self-induction. Coefficient of self-induction
The self-inductance of a coil is given by,
$e = - L\dfrac{{d\phi }}{{dt}}{\text{ }} \to 1$
The magnetic flux associated with the coil is proportional to the current, when a current flows in a coil,
$\Rightarrow \phi \propto I$
$\Rightarrow \phi = LI$
The equation 1 becomes
$e = - L\dfrac{{dI}}{{dt}}{\text{ }} \to 2$
Where,
$\phi$ is the magnetic flux
$d\phi$ is the change in magnetic flux
e is the emf induced in the coil.
dI is the change in the current.
dt is the change in the time.
L is a constant and it is known as the coefficient of self-inductance or self-induction.
The coefficient of self-induction of a coil is equal to the magnetic flux associated with a coil when unit current flows through it.
We can get the co-efficient of the self-inductance can be derived from the self-inductance formula,
$L = - \dfrac{e}{{\dfrac{{dI}}{{dt}}}}{\text{ }} \to 3$
The unit of self-inductance is henry (H).
The given statement is,
The electric current in a coil steadily changes from $+ 2{\text{A to - 2A}}$ in$0.05s$, an induced emf of $8.0V$ is generated in it.
The change in current $dI$ is $+ 2{\text{A to - 2A}}$
Initial current in the coil is $+ 2A$
The final current in the coil is $- 2A$
The change in current can be derived by subtracting the initial current from the final current.
$\Rightarrow dI = - 2A - ( + 2A)$
$\Rightarrow dI = - 4A$
The change in time, $dt = 0.05s$
The emf induced due to the change in the current, $e = 8.0V$
The co-efficient of self-inductance or the self-induction is
$L = - \dfrac{e}{{\dfrac{{dI}}{{dt}}}}$
Substitute the known as
$\Rightarrow L = - \dfrac{8}{{\dfrac{{ - 4}}{{0.05}}}}$
$\Rightarrow L = - \dfrac{8}{{ - 80}}$
$\Rightarrow L = + \dfrac{1}{{10}}$
$\Rightarrow L = 0.1H$

Hence the self-induction of the coil is $0.1H$

Note: Most of the students forget the negative sign in the self-induction formula. The emf induced in the coil opposes the current flow in the coil so we must use the negative sign. Students who forget to put the negative sign would get a wrong answer.