A rectangular coil of `20` turns and area of cross-section `25 cm^(2)` has a resistance of `100 ohm`. If a magnetic field which is perpendicular to the plane of the coil changes at the rate of `1000` tesla per second, the current in the coil is

Options

A. `1.0` ampere
B. `50` ampere
C. `0.5` ampere
D. `5.0` ampere

Explanation

To solve the problem step by step, we will use the principles of electromagnetic induction and Ohm's law.

### Step 1: Understand the given values
- Number of turns (N) = 20
- Area of the coil (A) = 25 cm² = 25 × 10⁻⁴ m² (since 1 cm² = 10⁻⁴ m²)
- Resistance (R) = 100 ohms
- Rate of change of magnetic field (dB/dt) = 1000 T/s

### Step 2: Calculate the magnetic flux (Φ)
The magnetic flux (Φ) through the coil is given by:
\[ \Phi = B \cdot A \cdot \cos(\theta) \]
Since the magnetic field is perpendicular to the plane of the coil, \(\theta = 0\) degrees, and \(\cos(0) = 1\).
Thus, the flux can be simplified to:
\[ \Phi = B \cdot A \]

### Step 3: Find the induced EMF (ε)
According to Faraday's law of electromagnetic induction, the induced EMF (ε) in the coil is given by the rate of change of magnetic flux:
\[ \epsilon = -\frac{d\Phi}{dt} \]
Substituting for Φ:
\[ \epsilon = -\frac{d(B \cdot A)}{dt} = -A \cdot \frac{dB}{dt} \]
Substituting the values:
\[ \epsilon = -25 \times 10^{-4} \, \text{m}^2 \cdot 1000 \, \text{T/s} \]
Calculating this gives:
\[ \epsilon = -25 \times 10^{-4} \times 1000 = -25 \times 10^{-1} = -2.5 \, \text{V} \]
(The negative sign indicates the direction of the induced EMF, but we will consider the magnitude for current calculation.)

### Step 4: Calculate the current (I)
Using Ohm's law, the current (I) can be calculated as:
\[ I = \frac{\epsilon}{R} \]
Substituting the values:
\[ I = \frac{2.5 \, \text{V}}{100 \, \Omega} = 0.025 \, \text{A} \]

### Step 5: Calculate the total current for the coil
Since the coil has 20 turns, the total current (I_total) through the coil will be:
\[ I_{\text{total}} = N \cdot I = 20 \cdot 0.025 \, \text{A} = 0.5 \, \text{A} \]

### Final Answer
The current in the coil is **0.5 A**.

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