A coil of N turns and mean cross-sectional area A is rotating with uniform angular velocity `omega` about an axis at right angle to uniform magnetic field B. The induced emf E in the coil will be

Options

A. `NBAsinomegat`
B. `NBomegasinomegat`
C. `NB//Asinomegat`
D. `NBAomegasinomegat`

Answer

To solve the problem of finding the induced EMF \( E \) in a rotating coil, we can follow these steps:

### Step 1: Understand the Setup
We have a coil with \( N \) turns and a mean cross-sectional area \( A \) rotating in a uniform magnetic field \( B \) with an angular velocity \( \omega \). The axis of rotation is perpendicular to the magnetic field.

### Step 2: Determine the Angle
As the coil rotates, the angle \( \theta \) between the area vector of the coil and the magnetic field changes over time. The angle can be expressed as:
\[ \theta(t) = \omega t \]
where \( \omega \) is the angular velocity and \( t \) is time.

### Step 3: Calculate Magnetic Flux
The magnetic flux \( \Phi \) through the coil at any instant is given by:
\[ \Phi = N \cdot B \cdot A \cdot \cos(\theta) \]
Substituting \( \theta(t) \):
\[ \Phi = N \cdot B \cdot A \cdot \cos(\omega t) \]

### Step 4: Find the Induced EMF
According to Faraday's law of electromagnetic induction, the induced EMF \( E \) is equal to the negative rate of change of magnetic flux:
\[ E = -\frac{d\Phi}{dt} \]
Now, we differentiate the expression for magnetic flux:
\[ E = -\frac{d}{dt}(N \cdot B \cdot A \cdot \cos(\omega t)) \]
Using the chain rule, we find:
\[ E = -N \cdot B \cdot A \cdot \frac{d}{dt}(\cos(\omega t)) \]
The derivative of \( \cos(\omega t) \) is:
\[ \frac{d}{dt}(\cos(\omega t)) = -\omega \sin(\omega t) \]
Thus, substituting this back, we have:
\[ E = -N \cdot B \cdot A \cdot (-\omega \sin(\omega t)) \]
This simplifies to:
\[ E = N \cdot B \cdot A \cdot \omega \sin(\omega t) \]

### Final Answer
The induced EMF \( E \) in the coil is given by:
\[ E = N \cdot B \cdot A \cdot \omega \sin(\omega t) \]

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