Question

An inductor is connected to a battery through a switch. Induced emf is \(e_1\) when the switch is pressed and \(e_2\) when the switch is opened. Then

• A. \(e_1 = e_2\)
• B. \(e_1>e_2\)
• C. \(e_1<e_2\)
• D. \(e_1>e_2\) or \(e_1<e_2\) depending on battery voltage

Hint:

Induced emf always opposes the change in current (Lenz's law).

Answer 1

The Correct Option is A

To solve this question, we need to understand the basic concept of electromagnetic induction, particularly as it pertains to inductors.

When the switch connected to an inductor is pressed (closed), a current starts flowing through the inductor, creating a magnetic field. According to Faraday's Law of Electromagnetic Induction, a changing magnetic field induces an electromotive force (emf) in the inductor. The direction of this induced emf is such that it opposes the change in current, as described by Lenz's Law.

Similarly, when the switch is opened, the current through the inductor drops to zero, causing another change in the magnetic field. This also induces an emf in the inductor, and just like before, the induced emf opposes the change in current.

The induced emf, \(e\), in an inductor is given by the formula:

\(e = -L \frac{dI}{dt}\)

where:

• \(e\) is the induced emf,
• \(L\) is the inductance of the inductor,
• \(\frac{dI}{dt}\) is the rate of change of current.

When the switch is pressed or opened, the rate of change of current \(\frac{dI}{dt}\) is the same in magnitude but opposite in direction, because the current is either starting from zero and increasing or dropping to zero and decreasing. Therefore, the magnitude of the induced emf when the switch is pressed (\(e_1\)) is equal to the magnitude of the induced emf when the switch is opened (\(e_2\)).

Thus, we conclude that \(e_1 = e_2\).

Therefore, the correct answer is: \(e_1 = e_2\).