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
C. 1.5
D. 2

Explanation

To solve the problem, we need to find the magnetic flux density (B) using the given data. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the relationship between charge, resistance, and magnetic flux
According to Faraday's law of electromagnetic induction, the induced electromotive force (EMF) in a coil is equal to the rate of change of magnetic flux through the coil. The induced EMF (ε) can also be expressed in terms of the charge (Q) that flows through the coil and the resistance (R) of the coil:
\[ \epsilon = \frac{Q}{R} \]

### Step 2: Relate the induced EMF to magnetic flux
The induced EMF can also be expressed in terms of the change in magnetic flux (ΔΦ) and the number of turns (N) in the coil:
\[ \epsilon = -N \frac{d\Phi}{dt} \]
Since the coil is suddenly removed from the magnetic field, the change in magnetic flux (ΔΦ) can be considered as the initial magnetic flux (Φ_initial) minus the final magnetic flux (Φ_final), which is zero:
\[ \Delta \Phi = N \cdot B \cdot A \]
Where:
- \(N\) = number of turns (40 turns)
- \(B\) = magnetic flux density (Wb/m²)
- \(A\) = area of the coil (in m²)

### Step 3: Convert the area from cm² to m²
The area is given as \(4.0 \, \text{cm}^2\). To convert this to square meters:
\[ A = 4.0 \, \text{cm}^2 = 4.0 \times 10^{-4} \, \text{m}^2 \]

### Step 4: Substitute the values into the equations
From the two expressions for induced EMF, we can set them equal to each other:
\[ \frac{Q}{R} = N \cdot \frac{\Delta \Phi}{dt} \]
Since the change in magnetic flux is \(N \cdot B \cdot A\), we can rewrite it as:
\[ \frac{Q}{R} = N \cdot B \cdot A \]

### Step 5: Solve for magnetic flux density (B)
Rearranging the equation to solve for \(B\):
\[ B = \frac{Q}{R \cdot N \cdot A} \]

### Step 6: Substitute the known values
Now we can substitute the known values into the equation:
- \(Q = 2.0 \times 10^{-4} \, \text{C}\)
- \(R = 80 \, \Omega\)
- \(N = 40\)
- \(A = 4.0 \times 10^{-4} \, \text{m}^2\)

Substituting these values:
\[ B = \frac{2.0 \times 10^{-4}}{80 \cdot 40 \cdot 4.0 \times 10^{-4}} \]

### Step 7: Calculate B
Calculating the denominator:
\[ 80 \cdot 40 \cdot 4.0 \times 10^{-4} = 128 \times 10^{-4} = 1.28 \times 10^{-2} \]
Now substituting back:
\[ B = \frac{2.0 \times 10^{-4}}{1.28 \times 10^{-2}} \approx 0.015625 \, \text{Wb/m}^2 \]

### Step 8: Final answer
Thus, the magnetic flux density \(B\) is approximately:
\[ B \approx 0.015625 \, \text{Wb/m}^2 \]