Question

The magnitude of flux linked with coil varies with time as $\phi = 3t^2 + 4t + 7$. The magnitude of induced e.m.f. at $t = 2\ \text{s}$ is

• A. 3 V
• B. 16 V
• C. 10 V
• D. 7 V

Hint:

Whenever flux is given as a polynomial function of time, the induced e.m.f. is simply its instantaneous rate of change. Differentiate first, then plug in the time value. Never plug the time value into the flux equation before differentiating, as that would wrongly yield a constant with a derivative of zero!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the magnitude of the electromotive force (e.m.f.) induced in a conducting coil at a specific instant ($t = 2\ \text{s}$), given the time-dependent equation for the magnetic flux ($\phi$) passing through it.

Step 2: Key Formula or Approach:
According to Faraday's Law of Electromagnetic Induction, the magnitude of the induced e.m.f. ($e$) is equal to the time derivative of the magnetic flux linkage:
$$e = \left| \frac{d\phi}{dt} \right|$$

Step 3: Detailed Explanation:
We are given the function for magnetic flux over time:
$$\phi = 3t^2 + 4t + 7$$ Differentiate this expression with respect to time $t$ using standard power rules ($\frac{d}{dt}[t^n] = nt^{n-1}$):
$$\frac{d\phi}{dt} = \frac{d}{dt}\left(3t^2 + 4t + 7\right)$$ $$\frac{d\phi}{dt} = (3 \times 2t) + 4 + 0 = 6t + 4$$ To find the instantaneous induced e.m.f. at the given timestamp, evaluate this derivative at $t = 2\ \text{s}$:
$$e = 6(2) + 4$$ $$e = 12 + 4 = 16\ \text{V}$$ This calculation gives an exact magnitude of 16 V for the induced e.m.f. at that instant.

Step 4: Final Answer:
The magnitude of the induced e.m.f. at $t = 2\ \text{s}$ is 16 V, which matches option (B).