A coil of resistance `10 Omega` and 1000 turns have the magnetic flux line of `5.5 xx10^(-4)`Wb. If the magnetic flux changed `5xx10^(-4)`Wb. In 0.1 sec, then the induced charge in coil is :-

Options

A. `50 mu C`
B. `5 mu C`
C. `2 muC`
D. `20 muC`

Answer

To solve the problem step by step, we will calculate the induced charge in the coil using the given data.

### Step 1: Calculate the change in magnetic flux (ΔΦ)
The initial magnetic flux (Φ_initial) is given as:
\[ Φ_{\text{initial}} = 5.5 \times 10^{-4} \, \text{Wb} \]
The final magnetic flux (Φ_final) is given as:
\[ Φ_{\text{final}} = 5.0 \times 10^{-4} \, \text{Wb} \]
The change in magnetic flux (ΔΦ) is calculated as:
\[ ΔΦ = Φ_{\text{final}} - Φ_{\text{initial}} = 5.0 \times 10^{-4} - 5.5 \times 10^{-4} = -0.5 \times 10^{-4} \, \text{Wb} \]

### Step 2: Calculate the induced EMF (E)
The induced EMF (E) can be calculated using Faraday's law of electromagnetic induction:
\[ E = -\frac{ΔΦ}{Δt} \]
Given that the time interval (Δt) is 0.1 seconds, we substitute the values:
\[ E = -\frac{-0.5 \times 10^{-4}}{0.1} = 5.0 \times 10^{-4} \, \text{V} \]

### Step 3: Calculate the induced current (I)
Using Ohm's law, the induced current (I) can be calculated as:
\[ I = \frac{E}{R} \]
Where R is the resistance of the coil, given as 10 Ω. Substituting the values:
\[ I = \frac{5.0 \times 10^{-4}}{10} = 5.0 \times 10^{-5} \, \text{A} \]

### Step 4: Calculate the induced charge (Q)
The induced charge (Q) can be calculated using the formula:
\[ Q = I \times Δt \]
Substituting the values we have:
\[ Q = 5.0 \times 10^{-5} \times 0.1 = 5.0 \times 10^{-6} \, \text{C} \]

### Step 5: Convert the charge to microcoulombs
To express the charge in microcoulombs (μC):
\[ Q = 5.0 \times 10^{-6} \, \text{C} = 5.0 \, \mu\text{C} \]

### Final Answer
The induced charge in the coil is:
\[ \boxed{5 \, \mu\text{C}} \]