If a coil of `40` turns and area `4.0 cm^(2)` is suddenly remove from a magnetic field, it is observed that a charge of `2.0xx10^(-4)C` flows into the coil. If the resistance of the coil is `80Omega`, the magnetic flux density in `Wb//m^(2)` is

Options

A. `0.5`
B. `1.0`
C. `1.5`
D. `2.0`

Explanation

To solve the problem, we need to find the magnetic flux density (B) in the coil when it is suddenly removed from the magnetic field. We can use the relationship between induced electromotive force (emf), charge, resistance, and magnetic flux.

### Step-by-Step Solution:

1. **Identify the given values:**
- Number of turns in the coil (N) = 40 turns
- Area of the coil (A) = 4.0 cm² = 4.0 × 10^(-4) m² (conversion from cm² to m²)
- Charge (Q) = 2.0 × 10^(-4) C
- Resistance (R) = 80 Ω

2. **Calculate the induced emf (E) using the formula:**
\[ E = \frac{Q}{t} \]
However, we do not have the time (t) directly. Instead, we can relate the induced emf to the resistance and charge:
\[ E = \frac{Q}{R} \]
Substituting the values:
\[ E = \frac{2.0 \times 10^{-4} \text{ C}}{80 \text{ Ω}} = 2.5 \times 10^{-6} \text{ V} \]

3. **Use Faraday's law of electromagnetic induction:**
The induced emf (E) is also given by:
\[ E = -N \frac{d\Phi}{dt} \]
Where \(\Phi\) is the magnetic flux. Since the coil is removed from the magnetic field, the change in magnetic flux (\(d\Phi\)) is equal to the initial magnetic flux (\(\Phi\)) because it goes from a certain value to zero.

4. **Calculate the magnetic flux (\(\Phi\)):**
Rearranging Faraday's law gives us:
\[ \Phi = -\frac{E \cdot dt}{N} \]
Since we are looking for the initial magnetic flux, we can consider \(dt\) as 1 second for simplicity (the exact time does not affect the calculation of B). Thus:
\[ \Phi = \frac{2.5 \times 10^{-6} \text{ V}}{40} = 6.25 \times 10^{-8} \text{ Wb} \]

5. **Calculate the magnetic flux density (B):**
The magnetic flux (\(\Phi\)) is related to the magnetic flux density (B) and the area (A) by the formula:
\[ \Phi = B \cdot A \]
Rearranging gives:
\[ B = \frac{\Phi}{A} \]
Substituting the values:
\[ B = \frac{6.25 \times 10^{-8} \text{ Wb}}{4.0 \times 10^{-4} \text{ m}^2} = 1.5625 \times 10^{-4} \text{ Wb/m}^2 \]

### Final Answer:
The magnetic flux density (B) is approximately:
\[ B \approx 1.56 \times 10^{-4} \text{ Wb/m}^2 \]