An emf of 25.0 mV is induced in a 500-turn coil when the current is changing is at the rate of 10.0A/s. What is the magnetic flux through each turn of the coil at an instant when the current is 4.0A?

Options

A. `20muT.m^2`
B. `10muT.m^2`
C. `15muT.m^2`
D. `30muT.m^2`

Answer

To solve the problem step by step, we will follow the provided information and apply the relevant formulas.

### Step 1: Identify the given values
- Induced EMF (ε) = 25.0 mV = 25 × 10^(-3) V
- Number of turns (N) = 500
- Rate of change of current (dI/dt) = 10.0 A/s
- Current at the instant (I) = 4.0 A

### Step 2: Use the formula for induced EMF
The induced EMF (ε) in a coil is related to its inductance (L) and the rate of change of current (dI/dt) by the formula:
\[ \epsilon = L \frac{dI}{dt} \]

### Step 3: Solve for inductance (L)
Rearranging the formula to find L:
\[ L = \frac{\epsilon}{\frac{dI}{dt}} \]

Substituting the known values:
\[ L = \frac{25 \times 10^{-3}}{10} = 2.5 \times 10^{-3} \, \text{H} \]
So, the inductance \( L \) is 2.5 mH.

### Step 4: Use the relationship between inductance, magnetic flux, and current
The relationship is given by:
\[ L = \frac{N \Phi}{I} \]
Where:
- \( \Phi \) is the magnetic flux through each turn.

Rearranging to find the magnetic flux \( \Phi \):
\[ \Phi = \frac{L \cdot I}{N} \]

### Step 5: Substitute the values to find magnetic flux
Now substituting the values we have:
\[ \Phi = \frac{(2.5 \times 10^{-3}) \cdot 4}{500} \]

Calculating:
\[ \Phi = \frac{10 \times 10^{-3}}{500} = \frac{10^{-2}}{500} = 20 \times 10^{-6} \, \text{Wb} \]
Thus, the magnetic flux through each turn of the coil is:
\[ \Phi = 20 \, \mu Wb \]

### Final Answer
The magnetic flux through each turn of the coil at the instant when the current is 4.0 A is \( 20 \, \mu Wb \).

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