Current (i) passing through a coil varies with time t as `i=2t^(2)`. At 1 s total flux passing through the coil is 10 Wb. Then

Options

A. self inductance of the coil is 10H
B. self inductance of the coil is 5H
C. induced emf across the coil at 1second is 20V
D. induced emf across the coil at 1second is 10V

Answer

To solve the problem step-by-step, we will follow these steps:

### Step 1: Determine the current at t = 1 second
Given the current \( i(t) = 2t^2 \), we can find the current at \( t = 1 \) second.

\[ i(1) = 2(1^2) = 2 \, \text{A} \]

### Step 2: Use the formula for self-inductance
The self-inductance \( L \) can be calculated using the formula:

\[ L = \frac{\Phi}{I} \]

where \( \Phi \) is the magnetic flux and \( I \) is the current. We know that at \( t = 1 \) second, the total flux \( \Phi = 10 \, \text{Wb} \) and the current \( I = 2 \, \text{A} \).

### Step 3: Calculate the self-inductance
Substituting the values into the formula:

\[ L = \frac{10 \, \text{Wb}}{2 \, \text{A}} = 5 \, \text{H} \]

### Step 4: Determine the induced EMF
The induced EMF \( E \) can be calculated using the formula:

\[ E = -L \frac{dI}{dt} \]

First, we need to find \( \frac{dI}{dt} \). Since \( I = 2t^2 \), we differentiate with respect to \( t \):

\[ \frac{dI}{dt} = \frac{d}{dt}(2t^2) = 4t \]

At \( t = 1 \) second:

\[ \frac{dI}{dt} = 4(1) = 4 \, \text{A/s} \]

### Step 5: Calculate the induced EMF at t = 1 second
Now substituting \( L = 5 \, \text{H} \) and \( \frac{dI}{dt} = 4 \, \text{A/s} \) into the EMF formula:

\[ E = -5 \times 4 = -20 \, \text{V} \]

The negative sign indicates the direction of the induced EMF according to Lenz's law, but the magnitude is what we are interested in, which is \( 20 \, \text{V} \).

### Summary of Results
- Self-inductance \( L = 5 \, \text{H} \)
- Induced EMF \( E = 20 \, \text{V} \)