A coil of `50` turns is pulled in `0.02s` between the poles of a magnet, where its area includes `31xx10^(-6)Wb` to `1xx10^(-6)Wb`.The average is

Options

A. `7.5xx10^(-2)V`
B. `7.5xx10^(-3)V`
C. zero
D. `7.5xx10^(-4)V`

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the given values
- Number of turns in the coil, \( n = 50 \)
- Initial magnetic flux, \( \Phi_i = 31 \times 10^{-6} \, \text{Wb} \)
- Final magnetic flux, \( \Phi_f = 1 \times 10^{-6} \, \text{Wb} \)
- Time interval, \( \Delta t = 0.02 \, \text{s} \)

### Step 2: Calculate the change in magnetic flux
The change in magnetic flux, \( \Delta \Phi \), is given by:
\[ \Delta \Phi = \Phi_f - \Phi_i \]
Substituting the values:
\[ \Delta \Phi = (1 \times 10^{-6}) - (31 \times 10^{-6}) = -30 \times 10^{-6} \, \text{Wb} \]

### Step 3: Use Faraday's law of electromagnetic induction
According to Faraday's law, the average induced EMF (\( E \)) is given by:
\[ E = -n \frac{\Delta \Phi}{\Delta t} \]
Substituting the values we have:
\[ E = -50 \times \frac{-30 \times 10^{-6}}{0.02} \]

### Step 4: Calculate the induced EMF
Now, we can simplify the expression:
\[ E = 50 \times \frac{30 \times 10^{-6}}{0.02} \]
Calculating the fraction:
\[ \frac{30 \times 10^{-6}}{0.02} = 1.5 \times 10^{-3} \]
Now multiply by 50:
\[ E = 50 \times 1.5 \times 10^{-3} = 75 \times 10^{-3} \, \text{V} = 0.075 \, \text{V} \]

### Final Answer
The average induced EMF is:
\[ E = 0.075 \, \text{V} \]

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