A coil having n turns and area A is initially placed with its plane normal to the magnetic field B. It is then rotated through `180^(@)` in 0.2 sec. The emf induced at the ends of the coils is ___

Options

A. `0.1 nAB`
B. nAB
C. 5 nAB
D. 10 nAB

Answer

To solve the problem, we need to calculate the induced electromotive force (emf) in a coil when it is rotated in a magnetic field. Here are the steps to find the solution:

### Step 1: Understand the Initial Conditions
The coil has:
- **n** turns
- **Area** \( A \)
- **Magnetic field** \( B \)
- Initially, the plane of the coil is normal (perpendicular) to the magnetic field.

### Step 2: Calculate the Initial Magnetic Flux
The magnetic flux \( \Phi_i \) through the coil when it is perpendicular to the magnetic field is given by:
\[ \Phi_i = n \cdot B \cdot A \cdot \cos(0^\circ) = n \cdot B \cdot A \cdot 1 = nBA \]

### Step 3: Determine the Final Conditions
After rotating the coil through \( 180^\circ \):
- The area vector of the coil is now opposite to the magnetic field direction.
- The angle between the magnetic field and the area vector is \( 180^\circ \).

### Step 4: Calculate the Final Magnetic Flux
The final magnetic flux \( \Phi_f \) is:
\[ \Phi_f = n \cdot B \cdot A \cdot \cos(180^\circ) = n \cdot B \cdot A \cdot (-1) = -nBA \]

### Step 5: Calculate the Change in Magnetic Flux
The change in magnetic flux \( \Delta \Phi \) is given by:
\[ \Delta \Phi = \Phi_f - \Phi_i = (-nBA) - (nBA) = -2nBA \]

### Step 6: Calculate the Induced EMF
According to Faraday's law of electromagnetic induction, the induced emf \( \mathcal{E} \) is given by the rate of change of magnetic flux:
\[ \mathcal{E} = -\frac{\Delta \Phi}{\Delta t} \]
where \( \Delta t = 0.2 \, \text{s} \). Thus,
\[ \mathcal{E} = -\frac{-2nBA}{0.2} = \frac{2nBA}{0.2} = 10nBA \]

### Final Answer
The induced emf at the ends of the coil is:
\[ \mathcal{E} = 10nAB \]

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