Question

The magnetic flux linked with a coil satisfies the relation $ \phi = (4t^2 + 6t + 9) \, \text{Wb} $, where $ t $ is time in seconds. The emf induced in the coil at $ t = 2 $ seconds is

• A. 22 V
• B. 18 V
• C. 16 V
• D. 40 V

Hint:

To calculate the induced emf, differentiate the magnetic flux \( \phi \) with respect to time \( t \) and evaluate it at the given time.

Answer 1

The Correct Option is A

To determine the emf induced in the coil, we use Faraday's law of electromagnetic induction, which states that the induced emf (\( \mathcal{E} \)) in a coil is equal to the negative rate of change of magnetic flux (\( \phi \)) through the coil. Mathematically, this is expressed as:

\[\mathcal{E} = -\frac{d\phi}{dt}\]

Given the magnetic flux linked with the coil:

\[\phi = 4t^2 + 6t + 9\]

we first calculate the derivative of \( \phi \) with respect to time \( t \):

\[\frac{d\phi}{dt} = \frac{d}{dt}(4t^2 + 6t + 9)\]

Applying basic differentiation rules, we get:

\[\frac{d\phi}{dt} = 8t + 6\]

Substituting \( t = 2 \) seconds into the derivative, we find:

\[\frac{d\phi}{dt}\bigg|_{t=2} = 8(2) + 6 = 16 + 6 = 22 \, \text{Wb/s}\]

Hence, the induced emf is:

\[\mathcal{E} = -\frac{d\phi}{dt}\bigg|_{t=2} = -22 \, \text{V}\]

Since the question asks only for the magnitude of the emf, the answer is 22 V.

Answer 2

The emf induced in a coil is given by Faraday's law of electromagnetic induction: \[ \mathcal{E} = -\frac{d\phi}{dt} \] where: 
- \( \mathcal{E} \) is the induced emf, 
- \( \phi \) is the magnetic flux. The magnetic flux is given by: \[ \phi = (4t^2 + 6t + 9) \, \text{Wb} \] To find the induced emf, we differentiate \( \phi \) with respect to \( t \): \[ \frac{d\phi}{dt} = \frac{d}{dt} \left( 4t^2 + 6t + 9 \right) \] Differentiating term by term: \[ \frac{d\phi}{dt} = 8t + 6 \] Now, substituting \( t = 2 \) seconds into the equation: \[ \frac{d\phi}{dt} = 8(2) + 6 = 16 + 6 = 22 \, \text{V} \] 
Therefore, the induced emf at \( t = 2 \) seconds is: \[ \mathcal{E} = 22 \, \text{V} \] 
Thus, the correct answer is: \( \text{(A) 22 V} \)