Two coils, held in fixed positions have a mutual inductance of 100`mu`H. what is the peak emf in one coil when the current in the other coil is 10 sin (1000t) where i is in amperes and t is in seconds?

Options

A. 2.00V
B. 1.00V
C. 4.00V
D. 3.00V

Answer

To solve the problem step by step, we will follow the principles of mutual inductance and differentiate the given current function.

### Step 1: Understand the given values
- Mutual inductance (M) = 100 µH = 100 × 10^(-6) H
- Current in the other coil: I(t) = 10 sin(1000t) A

### Step 2: Write the formula for induced EMF
The induced electromotive force (emf) in one coil due to the changing current in another coil is given by the formula:
\[ E = -M \frac{dI}{dt} \]
where:
- E is the induced emf,
- M is the mutual inductance,
- \(\frac{dI}{dt}\) is the rate of change of current with respect to time.

### Step 3: Differentiate the current function
We need to find \(\frac{dI}{dt}\) for the given current function:
\[ I(t) = 10 \sin(1000t) \]
Using the chain rule for differentiation:
\[ \frac{dI}{dt} = 10 \cdot \frac{d}{dt}(\sin(1000t)) \]
The derivative of \(\sin(1000t)\) is:
\[ \frac{d}{dt}(\sin(1000t)) = 1000 \cos(1000t) \]
Thus,
\[ \frac{dI}{dt} = 10 \cdot 1000 \cos(1000t) = 10000 \cos(1000t) \]

### Step 4: Substitute into the emf formula
Now substitute \(\frac{dI}{dt}\) back into the emf formula:
\[ E = -M \frac{dI}{dt} \]
\[ E = -100 \times 10^{-6} \cdot 10000 \cos(1000t) \]

### Step 5: Simplify the expression
Calculating the product:
\[ E = -100 \times 10^{-6} \cdot 10000 \cos(1000t) \]
\[ E = -1 \cos(1000t) \]

### Step 6: Find the peak value of the induced emf
The peak value of the induced emf, denoted as \(E_0\), is the coefficient of the cosine function:
\[ E_0 = 1 \text{ V} \]

### Final Answer
The peak emf in one coil is:
\[ E_0 = 1 \text{ V} \]

---