The magnetic flux linked with a coil (in Wb) is given by the equation `phi = 5t^2 + 3t +16` . The magnetic of induced emf in the coil at fourth second will be

Options

A. 33V
B. 43V
C. 108V
D. 10V

Answer

To find the magnitude of the induced EMF in the coil at the fourth second, we need to follow these steps:

### Step 1: Understand the relationship between magnetic flux and induced EMF
The induced EMF (ε) in a coil is given by Faraday's law of electromagnetic induction, which states that:
\[ \varepsilon = -\frac{d\Phi}{dt} \]
where \(\Phi\) is the magnetic flux.

### Step 2: Differentiate the magnetic flux equation
Given the magnetic flux linked with the coil is:
\[ \Phi = 5t^2 + 3t + 16 \]
We need to differentiate this equation with respect to time \(t\) to find \(\frac{d\Phi}{dt}\).

### Step 3: Perform the differentiation
Differentiating \(\Phi\):
\[ \frac{d\Phi}{dt} = \frac{d}{dt}(5t^2 + 3t + 16) \]
Using the power rule of differentiation:
\[ \frac{d\Phi}{dt} = 10t + 3 \]

### Step 4: Substitute \(t = 4\) seconds into the derivative
Now, we need to find the value of \(\frac{d\Phi}{dt}\) at \(t = 4\) seconds:
\[ \frac{d\Phi}{dt} = 10(4) + 3 = 40 + 3 = 43 \]

### Step 5: Calculate the induced EMF
Now, substituting this value into the equation for induced EMF:
\[ \varepsilon = -\frac{d\Phi}{dt} = -43 \]
Since we are asked for the magnitude of the induced EMF, we take the absolute value:
\[ |\varepsilon| = 43 \text{ volts} \]

### Final Answer
The magnitude of the induced EMF in the coil at the fourth second is **43 volts**.

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