The magnetic flux passing perpendicular to the plane of coil is changing according to the equation $\Phi = 6t^2+7t + 1$, where '$\Phi$' is in mWb and 't' in sec. The magnitude of induced emf at t = 2s is

31 mV

The correct answer is Option (1) → 31 mV ##

To calculate the magnitude of the induced electromotive force (emf), we use Faraday's Law of Electromagnetic Induction. This law states that the induced emf is equal to the negative rate of change of magnetic flux through the circuit.

Since the question asks for the magnitude, we can focus on the absolute value of the derivative:

$|\varepsilon| = \left| \frac{d\Phi}{dt} \right|$

1. Differentiate the Equation

The given equation for flux is $\Phi = 6t^2 + 7t + 1$. We differentiate this with respect to time ($t$):

$\frac{d\Phi}{dt} = \frac{d}{dt}(6t^2 + 7t + 1)$

$\frac{d\Phi}{dt} = 12t + 7$

2. Substitute the Time Value

The question asks for the magnitude of induced emf at $t = 2$ s. We substitute $t = 2$ into our derivative:

$|\varepsilon| = 12(2) + 7$

$|\varepsilon| = 24 + 7$

$|\varepsilon| = 31$

3. Determine the Units

Since the magnetic flux $\Phi$ is given in milliwebers (mWb) and time $t$ is in seconds (s), the resulting emf will be in millivolts (mV).

Result:

The magnitude of the induced emf at $t = 2$ s is 31 mV.