Question

A coil with a self-inductance of 7.9 H carries a current that decreases at a uniform rate dI/dt= 0.060 A/s. What is the emf induced in the coil (in V)?

Answer 1

The induced electromotive force (emf) in a coil with a self-inductance of 7.9 H when the current decreases at a rate of 0.060 A/s is 0.474 V.

Step-by-step explanation:

Calculating Induced EMF in a Coil

The question involves calculating the induced emf (electromotive force) in a coil due to a changing current. According to Faraday's law of electromagnetic induction, the magnitude of the induced emf (ε) in a coil is directly proportional to the rate of change of current through the coil and its self-inductance.

To find the emf induced in the coil, we use the formula:

ε = -L (dI/dt)

Where:

• L is the self-inductance of the coil (measured in henries, H)
• dI/dt is the rate of change of current through the coil (measured in amperes per second, A/s)

Given in the question:

• L = 7.9 H (self-inductance of the coil)
• dI/dt = 0.060 A/s (uniform rate at which the current decreases)

Substituting the given values into the formula, the induced emf is calculated as follows:

ε = -7.9 H * 0.060 A/s

ε = -0.474 V

The negative sign indicates that the direction of induced emf is such that it opposes the change in current through the coil, in accordance with Lenz's law. However, since the question asks for the magnitude of the induced emf, we report it as 0.474 V without the negative sign.

Therefore, the induced emf in the coil is 0.474 V.