A jet plane having a wing-span of 25 m is travelling horizontally towards East with a speed of 3600 km /hr. If the Earth's magnetic field at the location is `4 ×× 10 ^(-4)` T and the angle of dip is `30^@` , then the potential difference between the ends of the wing is

Options

A. 4 V
B. 5 V
C. 2 V
D. 2.5 V

Answer

To solve the problem, we need to calculate the potential difference (emf) induced between the ends of the wings of the jet plane as it travels through the Earth's magnetic field. Here are the steps to find the solution:

### Step 1: Convert the speed of the jet plane from km/hr to m/s
The speed of the jet plane is given as 3600 km/hr. We can convert this to meters per second using the conversion factor \(1 \text{ km/hr} = \frac{1}{3.6} \text{ m/s}\).

\[ \text{Speed in m/s} = \frac{3600 \text{ km/hr}}{3.6} = 1000 \text{ m/s} \]

### Step 2: Determine the vertical component of the Earth's magnetic field
The Earth's magnetic field \(B\) is given as \(4 \times 10^{-4} \text{ T}\) and the angle of dip \(\phi\) is \(30^\circ\). The vertical component \(B_v\) can be calculated using the formula:

\[ B_v = B \sin(\phi) \]

Substituting the values:

\[ B_v = 4 \times 10^{-4} \text{ T} \times \sin(30^\circ) = 4 \times 10^{-4} \times \frac{1}{2} = 2 \times 10^{-4} \text{ T} \]

### Step 3: Calculate the induced emf (potential difference)
The formula for the induced emf (\(E\)) in a conductor moving through a magnetic field is given by:

\[ E = B_v \cdot v \cdot L \]

Where:
- \(B_v\) is the vertical component of the magnetic field,
- \(v\) is the speed of the plane,
- \(L\) is the wingspan of the plane.

Substituting the known values:

\[ E = (2 \times 10^{-4} \text{ T}) \cdot (1000 \text{ m/s}) \cdot (25 \text{ m}) \]

Calculating this gives:

\[ E = 2 \times 10^{-4} \cdot 1000 \cdot 25 = 2 \times 25 = 50 \text{ V} \]

### Conclusion
The potential difference between the ends of the wing is \(50 \text{ V}\).