A 50 turns circular coil has a radius of 3 cm, it is kept in a magnetic field acting normal to the area of the coil. The magnetic field B increased from 0.10 tesla to 0.35 tesla in 2 milliseconds. The average induced e.m.f in the coil is

Options

A. 1.77 volts
B. 17.7 volts
C. 177 volts
D. 0.177 volts

Answer

To solve the problem of finding the average induced e.m.f in a circular coil placed in a magnetic field, we can follow these steps:

### Step 1: Identify the given values
- Number of turns (N) = 50
- Radius of the coil (r) = 3 cm = 3 × 10^(-2) m
- Initial magnetic field (B_i) = 0.10 T
- Final magnetic field (B_f) = 0.35 T
- Time interval (Δt) = 2 ms = 2 × 10^(-3) s

### Step 2: Calculate the area of the coil
The area (A) of a circular coil is given by the formula:
\[ A = \pi r^2 \]
Substituting the radius:
\[ A = \pi (3 \times 10^{-2})^2 \]
\[ A = \pi (9 \times 10^{-4}) \]
\[ A \approx 2.826 \times 10^{-3} \, \text{m}^2 \]

### Step 3: Calculate the change in magnetic flux
The magnetic flux (Φ) through the coil is given by:
\[ \Phi = N \cdot B \cdot A \]
The change in magnetic flux (ΔΦ) is:
\[ \Delta \Phi = \Phi_f - \Phi_i \]
Where:
- \( \Phi_f = N \cdot B_f \cdot A \)
- \( \Phi_i = N \cdot B_i \cdot A \)

Calculating each:
\[ \Phi_f = 50 \cdot 0.35 \cdot 2.826 \times 10^{-3} \]
\[ \Phi_f \approx 0.0495 \, \text{Wb} \]

\[ \Phi_i = 50 \cdot 0.10 \cdot 2.826 \times 10^{-3} \]
\[ \Phi_i \approx 0.0141 \, \text{Wb} \]

Now, calculate ΔΦ:
\[ \Delta \Phi = \Phi_f - \Phi_i \]
\[ \Delta \Phi \approx 0.0495 - 0.0141 \]
\[ \Delta \Phi \approx 0.0354 \, \text{Wb} \]

### Step 4: Calculate the average induced e.m.f (E)
The average induced e.m.f (E) is given by:
\[ E = -\frac{\Delta \Phi}{\Delta t} \]
Substituting the values:
\[ E = -\frac{0.0354}{2 \times 10^{-3}} \]
\[ E \approx -17.7 \, \text{V} \]

Since we are interested in the magnitude:
\[ E \approx 17.7 \, \text{V} \]

### Conclusion
The average induced e.m.f in the coil is approximately **17.7 volts**.

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