A flat coil of 500 turns each of area `50 cm^(2)` rotates in a uniform magnetic field of `0.14 Wb//m^(2)` at an angular speed of `150 rad//sec`. The coil has a resistance of `5 Omega`. The induced e.m.f. is applied to an external resistance of 10 ohm. Calculate the peak current through the resistance.

Options

A. 1.5 A
B. 2.5 A
C. 3.5 A
D. 4.5 A

Answer

To solve the problem step by step, we will follow the concepts of electromagnetic induction and use the relevant formulas.

### Step 1: Understand the Problem
We have a flat coil with:
- Number of turns (N) = 500
- Area of the coil (A) = 50 cm² = 50 × 10^(-4) m²
- Magnetic field (B) = 0.14 Wb/m²
- Angular speed (ω) = 150 rad/s
- Resistance of the coil (R_coil) = 5 Ω
- External resistance (R_ext) = 10 Ω

We need to find the peak current through the external resistance.

### Step 2: Calculate the Total Resistance
The total resistance (R_total) in the circuit is the sum of the resistance of the coil and the external resistance:
\[ R_{\text{total}} = R_{\text{coil}} + R_{\text{ext}} = 5 \, \Omega + 10 \, \Omega = 15 \, \Omega \]

### Step 3: Calculate the Magnetic Flux
The magnetic flux (Φ) through the coil is given by:
\[ \Phi = N \cdot B \cdot A \cdot \cos(\theta) \]
Since the coil is rotating, we can express θ as ωt. The time-dependent magnetic flux becomes:
\[ \Phi(t) = N \cdot B \cdot A \cdot \cos(\omega t) \]

### Step 4: Calculate the Induced EMF
The induced EMF (ε) in the coil can be calculated using Faraday's law of electromagnetic induction:
\[ \epsilon = -\frac{d\Phi}{dt} \]
Differentiating the magnetic flux:
\[ \epsilon = -\frac{d}{dt}(N \cdot B \cdot A \cdot \cos(\omega t)) = N \cdot B \cdot A \cdot \omega \cdot \sin(\omega t) \]

### Step 5: Calculate the Peak EMF
The peak value of the induced EMF (ε_peak) occurs when sin(ωt) = 1:
\[ \epsilon_{\text{peak}} = N \cdot B \cdot A \cdot \omega \]

### Step 6: Substitute the Values
Now substituting the values into the equation:
- N = 500
- B = 0.14 Wb/m²
- A = 50 × 10^(-4) m²
- ω = 150 rad/s

Calculating ε_peak:
\[ \epsilon_{\text{peak}} = 500 \cdot 0.14 \cdot (50 \times 10^{-4}) \cdot 150 \]
\[ = 500 \cdot 0.14 \cdot 0.005 \cdot 150 \]
\[ = 500 \cdot 0.14 \cdot 0.75 \]
\[ = 52.5 \, \text{V} \]

### Step 7: Calculate the Peak Current
Using Ohm's law, the peak current (I_peak) through the external resistance can be calculated as:
\[ I_{\text{peak}} = \frac{\epsilon_{\text{peak}}}{R_{\text{total}}} \]
Substituting the values:
\[ I_{\text{peak}} = \frac{52.5}{15} \]
\[ = 3.5 \, \text{A} \]

### Final Answer
The peak current through the resistance is **3.5 A**.

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