Question

The magnetic flux linked with a coil changes with time \( t \) as \( \phi = (8t^2 + 5t + 7) \), where \( t \) is in seconds and \( \phi \) is in Wb. The value of emf induced in the coil at \( t = 4 \) s is:

• A. 32 V
• B. 37 V
• C. 64 V
• D. 69 V

Hint:

To find the induced emf, differentiate the magnetic flux with respect to time. The induced emf is equal to the rate of change of magnetic flux.

Answer 1

The Correct Option is D

The induced emf in a coil is given by Faraday's law of electromagnetic induction, which states that: \[ \text{emf} = -\frac{d\phi}{dt} \] Where: - \( \phi \) is the magnetic flux, - \( \frac{d\phi}{dt} \) is the rate of change of flux. The given magnetic flux is: \[ \phi = 8t^2 + 5t + 7 \] To find the induced emf, we need to differentiate the flux with respect to time \( t \): \[ \frac{d\phi}{dt} = \frac{d}{dt} (8t^2 + 5t + 7) \] Differentiating each term: \[ \frac{d\phi}{dt} = 16t + 5 \] Now, substitute \( t = 4 \) seconds into this expression to find the induced emf at that time: \[ \frac{d\phi}{dt} = 16(4) + 5 = 64 + 5 = 69 \, \text{V} \] Thus, the induced emf in the coil at \( t = 4 \) s is 69 V. Therefore, the correct answer is option (D).