Question

The magnetic flux (in weber) linked with a coil of resistance 10 Ω is varying with respect to time \( t \) as \( \phi = 4t^{2} + 2t + 1 \). Then the current in the coil at time \( t = 1 \) second is

• A. 0.5 A
• B. 2 A
• C. 1.5 A
• D. 1 A
• E. 2.5 A

Hint:

Remember that emf is the \textbf{rate of change} of flux. If flux is a function of time, you must differentiate it before plugging in the specific time value.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
According to Faraday's law of electromagnetic induction, a time-varying magnetic flux through a coil induces an electromotive force (EMF) in the coil. This induced EMF then drives a current through the coil, which is governed by Ohm's law.
Step 2: Key Formula or Approach:
1. The magnitude of induced EMF (\( |\epsilon| \)) is the rate of change of magnetic flux: \[ |\epsilon| = \left| \frac{d\phi}{dt} \right| \] 2. The induced current (\( I \)) is given by Ohm's Law: \[ I = \frac{|\epsilon|}{R} \] where \( R \) is the resistance of the coil.
Step 3: Detailed Explanation:
The given equation for magnetic flux is: \[ \phi = 4t^2 + 2t + 1 \]
Differentiate the flux equation with respect to time \( t \) to find the induced EMF: \[ \frac{d\phi}{dt} = \frac{d}{dt}(4t^2 + 2t + 1) \] \[ \frac{d\phi}{dt} = 8t + 2 \]
Now, calculate the magnitude of the EMF at the specific instant \( t = 1 \) second: \[ |\epsilon| = |8(1) + 2| \] \[ |\epsilon| = 10 \text{ V} \]
Given the resistance of the coil is \( R = 10 \) (ohms), we calculate the induced current: \[ I = \frac{|\epsilon|}{R} \] \[ I = \frac{10}{10} \] \[ I = 1 \text{ A} \]
Step 4: Final Answer:
The current in the coil at time \( t = 1 \) second is \( 1 \text{ A} \).