A rectangular coil `20 cm xx 10 cm` having 500 turns rotates in a magnetic field of `5 xx 10^(-3) T` with a frequency of `1200 rev min^(-1)` about an axis perpendicular to the field.

Options

A. The maximum value of the induced emf `2pi//5` volt.
B. The instantaneous emf when the plane of the coil is perpendicular to the field is zero
C. The instantaneous emf when the plane of the coil makes an angle of `60^(@)` with the field is `pi//5` volt.
D. The instantaneous emf when the plane of the coil makes an angle of `30^(@)` with the field is `pi//10` volt

Explanation

To solve the problem step by step, we will calculate the induced EMF in the rectangular coil rotating in a magnetic field.

### Step 1: Calculate the Area of the Coil
The area \( A \) of the rectangular coil can be calculated using the formula:
\[ A = \text{length} \times \text{width} \]
Given dimensions:
- Length = 20 cm = 0.2 m
- Width = 10 cm = 0.1 m

Thus,
\[ A = 0.2 \, \text{m} \times 0.1 \, \text{m} = 0.02 \, \text{m}^2 \]

### Step 2: Convert Frequency to Radians per Second
The frequency \( f \) is given as 1200 revolutions per minute. To convert this to revolutions per second:
\[ f = \frac{1200 \, \text{rev/min}}{60 \, \text{s/min}} = 20 \, \text{rev/s} \]
Now, convert revolutions per second to radians per second using:
\[ \omega = 2\pi f \]
Thus,
\[ \omega = 2\pi \times 20 = 40\pi \, \text{rad/s} \]

### Step 3: Calculate the Maximum Induced EMF
The induced EMF \( E \) can be calculated using Faraday's law of electromagnetic induction:
\[ E = -N \frac{d\Phi}{dt} \]
Where \( \Phi \) (magnetic flux) is given by:
\[ \Phi = B \cdot A \cdot \cos(\theta) \]
Here, \( N \) is the number of turns (500), \( B \) is the magnetic field strength (5 x \( 10^{-3} \, \text{T} \)), and \( \theta \) is the angle between the magnetic field and the normal to the coil.

Since the coil is rotating, we can express \( \theta \) as \( \omega t \):
\[ \Phi = B \cdot A \cdot \cos(\omega t) \]

Taking the derivative:
\[ \frac{d\Phi}{dt} = -B \cdot A \cdot \omega \sin(\omega t) \]
Thus, the induced EMF becomes:
\[ E = N \cdot B \cdot A \cdot \omega \cdot \sin(\omega t) \]

### Step 4: Calculate the Maximum Value of EMF
The maximum induced EMF occurs when \( \sin(\omega t) = 1 \):
\[ E_{\text{max}} = N \cdot B \cdot A \cdot \omega \]
Substituting the values:
- \( N = 500 \)
- \( B = 5 \times 10^{-3} \, \text{T} \)
- \( A = 0.02 \, \text{m}^2 \)
- \( \omega = 40\pi \, \text{rad/s} \)

Calculating:
\[ E_{\text{max}} = 500 \cdot (5 \times 10^{-3}) \cdot (0.02) \cdot (40\pi) \]
\[ E_{\text{max}} = 500 \cdot 5 \cdot 0.02 \cdot 40\pi \times 10^{-3} \]
\[ E_{\text{max}} = 100 \cdot 40\pi \times 10^{-3} = 4\pi \, \text{V} \]

### Step 5: Evaluate Specific Conditions
1. **When the plane of the coil is perpendicular to the magnetic field**: The angle \( \theta = 90^\circ \) means \( \sin(\omega t) = 0 \), hence \( E = 0 \).
2. **When the plane of the coil makes a 60-degree angle with the magnetic field**: The angle between the magnetic field and the area vector is \( 30^\circ \) (since \( 90^\circ - 60^\circ = 30^\circ \)):
\[ E = N \cdot B \cdot A \cdot \omega \cdot \sin(30^\circ) = \frac{1}{2} N \cdot B \cdot A \cdot \omega \]
Substituting the values gives:
\[ E = \frac{1}{2} \cdot 4\pi = 2\pi \, \text{V} \]

### Step 6: When the plane of the coil makes a 30-degree angle with the field
The angle between the magnetic field and the area vector is \( 60^\circ \):
\[ E = N \cdot B \cdot A \cdot \omega \cdot \sin(60^\circ) = \frac{\sqrt{3}}{2} N \cdot B \cdot A \cdot \omega \]
Substituting the values gives:
\[ E = \frac{\sqrt{3}}{2} \cdot 4\pi = 2\sqrt{3}\pi \, \text{V} \]

### Summary of Results
- Maximum induced EMF: \( 4\pi \, \text{V} \)
- Induced EMF when perpendicular to the field: \( 0 \, \text{V} \)
- Induced EMF at 60 degrees: \( 2\pi \, \text{V} \)
- Induced EMF at 30 degrees: \( 2\sqrt{3}\pi \, \text{V} \)