The magnetic flux linked with a coil is `phi = (3 t^(2) - 2 t + 1)` milliweber. The e.m.f. induced in the coil at t = 1 sec is

Options

A. 4 V
B. `4 xx 10^(-3) V`
C. 6 V
D. `4 xx 10^(3) V`

Answer

To find the induced electromotive force (e.m.f.) in the coil at \( t = 1 \) second, we can follow these steps:

### Step 1: Write down the expression for magnetic flux
The magnetic flux linked with the coil is given by:
\[ \phi(t) = 3t^2 - 2t + 1 \text{ milliweber} \]

### Step 2: Convert milliweber to weber
Since we need to work in standard units, we convert milliweber to weber:
\[ \phi(t) = (3t^2 - 2t + 1) \times 10^{-3} \text{ weber} \]

### Step 3: Differentiate the magnetic flux with respect to time
To find the induced e.m.f. (\( E \)), we use Faraday's law of electromagnetic induction, which states:
\[ E = -\frac{d\phi}{dt} \]
Now, we differentiate \( \phi(t) \):
\[ \frac{d\phi}{dt} = \frac{d}{dt} \left( 3t^2 - 2t + 1 \right) \times 10^{-3} \]

Calculating the derivative:
\[ \frac{d\phi}{dt} = (6t - 2) \times 10^{-3} \]

### Step 4: Substitute \( t = 1 \) second into the derivative
Now, we substitute \( t = 1 \) into the derivative to find the induced e.m.f.:
\[ \frac{d\phi}{dt} \bigg|_{t=1} = (6(1) - 2) \times 10^{-3} = (6 - 2) \times 10^{-3} = 4 \times 10^{-3} \text{ weber/second} \]

### Step 5: Calculate the induced e.m.f.
Now, substituting back into the formula for e.m.f.:
\[ E = -\frac{d\phi}{dt} = - (4 \times 10^{-3}) \text{ volts} \]
Since e.m.f. is typically expressed as a positive quantity, we take the absolute value:
\[ E = 4 \times 10^{-3} \text{ volts} \]

### Final Answer
The induced e.m.f. in the coil at \( t = 1 \) second is:
\[ E = 4 \text{ mV} \text{ (or } 4 \times 10^{-3} \text{ volts)} \]

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