Question

A coil of inductance 0.20 H is connected in series with a switch and cell of emf 1.6V. The total resistance of the circuit is $4\Omega $. What is the initial rate of growth of the current when the switch is closed?
A) $0.050A{s^{ - 1}}$
B) $0.40A{s^{ - 1}}$
C) $0.13A{s^{ - 1}}$
D) $8.0A{s^{ - 1}}$

Answer 1

Hint: To solve this question, we have to understand the concept of back emf in an inductor. The back emf is an emf produced in the inductor when there is flow of an alternating current through the inductor, which is a result of the phenomenon of electromagnetic induction, based on Faraday's laws.

Complete step by step answer:
An inductor is a passive electrical device which consists of a coiled wire with several turns wound over a material known as core. The inductor behaves like a normal conductor when direct current flows through it but it offers obstruction to alternating current, which is termed as reactance.
The obstruction offered by the inductor is because of the existence of an opposing emf in the inductor which tends to oppose the alternating current flowing through it. This emf is called back emf.
The back emf exists because of the Faraday’s law which states that –
When magnetic flux linked with the coil changes, an emf is induced across its ends.
$\varepsilon = - \dfrac{{d\phi }}{{dt}}$
Now, when alternating current flows through the inductor, the magnetic flux linked with the inductor changes with the current. This results in an emf in the inductor known as the back emf.
Since, magnetic flux linked with the coil is proportional to the current through the coil, we have –
Back emf,
$v = - L\dfrac{{di}}{{dt}}$
where L = constant known as the coefficient of self-inductance (or simply, inductance) of the inductor measured in henry (H) and $\dfrac{{di}}{{dt}}$ = rate of change of current through the coil.
Given,
Inductance, $L = 0.20H$
Emf of the cell, $E = 1.6V$
Resistance in the circuit, $R = 4\Omega $
When the switch is closed, the current will take some time to set up in the circuit since the inductor causes a lag in the current in the circuit. Thus, at the point when the switch is closed, the current in the circuit is zero but the inductor is active.
Thus, the back emf of the inductor will be equal to the emf of the cell.
Hence, we have –
$v = E$
This gives us the relation –
$E = L\dfrac{{di}}{{dt}}$
Substituting the values, we get –
$\Rightarrow 1.6 = 0.2\dfrac{{di}}{{dt}}$
$ \Rightarrow \dfrac{{di}}{{dt}} = \dfrac{{1.6}}{{0.2}} = 8A{s^{ - 1}}$
Hence, the rate of change of current in the circuit is 8 amperes per second.

Hence, the correct option is Option D.

Note: The emf is in the negative direction because of the Lenz’ law which states that the emf generated due to the electromagnetic induction produces a magnetic field which tends to oppose the magnetic field that is causing it. This law is in accordance with the principal of conservation of energy. Hence, the emf induced will be in the negative direction.