In a magnetic field of `0.05T`, area of a coil changes from `101 cm^(2)` to `100 cm^(2)` without changing the resistance which is `2Omega`. The amount of charge that flow during this period is

Options

A. `2.5xx10^(-6)C`
B. `2xx10^(-6)C`
C. `10^(-6)C`
D. `8xx10^(-6)C`

Explanation

To solve the problem, we need to calculate the amount of charge that flows through a coil when its area changes in a magnetic field. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Magnetic field (B) = 0.05 T
- Initial area (A_initial) = 101 cm² = 101 × 10^(-4) m²
- Final area (A_final) = 100 cm² = 100 × 10^(-4) m²
- Resistance (R) = 2 Ω

### Step 2: Calculate the change in magnetic flux (ΔΦ)
The magnetic flux (Φ) through the coil is given by:
\[ \Phi = B \times A \]

The initial magnetic flux (Φ_initial) is:
\[ \Phi_{initial} = B \times A_{initial} = 0.05 \, \text{T} \times (101 \times 10^{-4} \, \text{m}^2) = 0.05 \times 1.01 \times 10^{-2} = 5.05 \times 10^{-4} \, \text{Wb} \]

The final magnetic flux (Φ_final) is:
\[ \Phi_{final} = B \times A_{final} = 0.05 \, \text{T} \times (100 \times 10^{-4} \, \text{m}^2) = 0.05 \times 1.00 \times 10^{-2} = 5.00 \times 10^{-4} \, \text{Wb} \]

The change in magnetic flux (ΔΦ) is:
\[ \Delta \Phi = \Phi_{final} - \Phi_{initial} = 5.00 \times 10^{-4} \, \text{Wb} - 5.05 \times 10^{-4} \, \text{Wb} = -0.05 \times 10^{-4} \, \text{Wb} = -5.0 \times 10^{-6} \, \text{Wb} \]

### Step 3: Calculate the induced EMF (ε)
Using Faraday's law of electromagnetic induction, the induced EMF (ε) is given by:
\[ \varepsilon = -\frac{\Delta \Phi}{\Delta t} \]

Since we don't have the time (Δt), we will keep it as a variable for now.

### Step 4: Relate EMF to current (I)
According to Ohm's law:
\[ I = \frac{\varepsilon}{R} \]

### Step 5: Relate current to charge (Q)
We know that:
\[ I = \frac{Q}{\Delta t} \]
Thus, we can express charge (Q) as:
\[ Q = I \cdot \Delta t = \frac{\varepsilon \cdot \Delta t}{R} \]

### Step 6: Substitute EMF into the charge equation
Substituting for ε:
\[ Q = \frac{-\Delta \Phi \cdot \Delta t}{R} \]

### Step 7: Substitute the values
Now substituting ΔΦ and R:
\[ Q = \frac{-(-5.0 \times 10^{-6}) \cdot \Delta t}{2} \]
\[ Q = \frac{5.0 \times 10^{-6} \cdot \Delta t}{2} \]

### Step 8: Calculate the charge
Since we are not given Δt, we can assume Δt = 1 second for simplicity, leading to:
\[ Q = \frac{5.0 \times 10^{-6}}{2} = 2.5 \times 10^{-6} \, \text{C} \]

### Final Answer
The amount of charge that flows during this period is:
\[ Q = 2.5 \times 10^{-6} \, \text{C} \]