A coil formed by wrapping 50 turns of wire in the shape of a square is positioned in a magnetic field so that the normal to the plane of the coil makes an angle of `30^@`. with the direction of the field. When the magnetic field is increased uniformly from `200 muT` to `600muT` in `0.4 s`, an emf of magnitude 80.0 mV is induced in the coil. What is the total length of the wire?

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We have a coil with 50 turns, shaped as a square, placed in a magnetic field. The magnetic field changes from 200 µT to 600 µT in 0.4 seconds, inducing an emf of 80.0 mV in the coil. We need to find the total length of the wire used in the coil.

### Step 2: Calculate the Change in Magnetic Field
The change in magnetic field (ΔB) can be calculated as:
\[ \Delta B = B_f - B_i = 600 \, \mu T - 200 \, \mu T = 400 \, \mu T = 400 \times 10^{-6} \, T \]

### Step 3: Calculate the Rate of Change of Magnetic Field
The rate of change of the magnetic field (dB/dt) is given by:
\[ \frac{dB}{dt} = \frac{\Delta B}{\Delta t} = \frac{400 \times 10^{-6} \, T}{0.4 \, s} = 1000 \times 10^{-6} \, T/s = 1 \, mT/s \]

### Step 4: Calculate the Induced EMF
The induced emf (ε) in the coil can be expressed as:
\[ \epsilon = -N \frac{d\Phi}{dt} \]
Where \( N \) is the number of turns and \( \Phi \) is the magnetic flux. The magnetic flux (Φ) is given by:
\[ \Phi = B \cdot A \cdot \cos(\theta) \]
Here, \( A \) is the area of the coil and \( \theta \) is the angle between the magnetic field and the normal to the coil (30 degrees).

### Step 5: Relate Induced EMF to Area
From the induced emf equation:
\[ \epsilon = N \cdot A \cdot \cos(30^\circ) \cdot \frac{dB}{dt} \]
Substituting the known values:
\[ 80 \times 10^{-3} = 50 \cdot A \cdot \cos(30^\circ) \cdot 1 \times 10^{-3} \]
Using \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \):
\[ 80 \times 10^{-3} = 50 \cdot A \cdot \frac{\sqrt{3}}{2} \cdot 1 \times 10^{-3} \]

### Step 6: Solve for Area (A)
Rearranging the equation gives:
\[ A = \frac{80 \times 10^{-3}}{50 \cdot \frac{\sqrt{3}}{2} \cdot 1 \times 10^{-3}} = \frac{80}{50 \cdot \frac{\sqrt{3}}{2}} = \frac{80 \cdot 2}{50 \sqrt{3}} = \frac{160}{50 \sqrt{3}} = \frac{16}{5 \sqrt{3}} \, m^2 \]

### Step 7: Calculate Side Length of the Square
Since the area of the square is \( A = s^2 \):
\[ s = \sqrt{A} = \sqrt{\frac{16}{5 \sqrt{3}}} \]
Calculating this gives:
\[ s \approx 1.36 \, m \]

### Step 8: Calculate the Total Length of the Wire
The perimeter (P) of the square is given by:
\[ P = 4s \]
Thus, the total length of the wire (L) for 50 turns is:
\[ L = 50 \cdot P = 50 \cdot 4s = 200s \]
Substituting the value of \( s \):
\[ L = 200 \cdot 1.36 \approx 272 \, m \]

### Final Answer
The total length of the wire is approximately **272 meters**.

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