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`

Answer

To solve the problem, we need to determine the magnetic flux density (B) in the coil when it is removed from the magnetic field. We will use the principles of electromagnetic induction and the relationship between induced EMF, current, charge, and resistance.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Number of turns (N) = 40
- Area of the coil (A) = 4.0 cm² = 4.0 × 10^(-4) m² (convert to square meters)
- Charge (Q) = 2.0 × 10^(-4) C
- Resistance (R) = 80 Ω

2. **Use the Formula for Induced EMF (ε):**
The induced EMF (ε) can be expressed in terms of the change in magnetic flux (ΔΦ) and the time taken (Δt):
\[ \epsilon = -\frac{\Delta \Phi}{\Delta t} \]

3. **Relate Current (I) to Induced EMF:**
The current (I) flowing through the coil can be defined as:
\[ I = \frac{\epsilon}{R} \]

4. **Relate Charge (Q) to Current:**
The charge (Q) that flows through the coil can also be expressed as:
\[ Q = I \cdot \Delta t \]

5. **Combine the Equations:**
From the above equations, we can write:
\[ Q = \frac{\epsilon}{R} \cdot \Delta t \]
Substituting for ε gives:
\[ Q = \frac{\Delta \Phi}{R \cdot \Delta t} \cdot \Delta t \]
This simplifies to:
\[ Q = \frac{\Delta \Phi}{R} \]

6. **Express Change in Magnetic Flux (ΔΦ):**
The change in magnetic flux (ΔΦ) when the coil is removed from the magnetic field can be expressed as:
\[ \Delta \Phi = N \cdot A \cdot B \]
where B is the magnetic flux density.

7. **Substituting Values:**
We can substitute the values we have into the equation:
\[ Q = \frac{N \cdot A \cdot B}{R} \]
Rearranging for B gives:
\[ B = \frac{Q \cdot R}{N \cdot A} \]

8. **Plug in the Values:**
Now substituting the known values:
\[ B = \frac{(2.0 \times 10^{-4} \, \text{C}) \cdot (80 \, \Omega)}{40 \cdot (4.0 \times 10^{-4} \, \text{m}^2)} \]

9. **Calculate B:**
- Calculate the numerator:
\[ 2.0 \times 10^{-4} \cdot 80 = 1.6 \times 10^{-2} \]
- Calculate the denominator:
\[ 40 \cdot 4.0 \times 10^{-4} = 1.6 \times 10^{-2} \]
- Thus:
\[ B = \frac{1.6 \times 10^{-2}}{1.6 \times 10^{-2}} = 1 \, \text{Wb/m}^2 \]

### Final Answer:
The magnetic flux density (B) is **1 Wb/m²**.

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