A boat is moving due east in a region where the earth's magnetic field is `5.0xx10^(-5) NA^(-1)m^(-1)` due north and horizontal. The boat carries a vertical aerial 2 m long. If the speed of the boat is `1.50 ms^(-1)`, the magnitude of the induced emf in the wire is

Options

A. 1 mV
B. 0.75 mV
C. 0.50 mV
D. 0.15mv

Explanation

To find the magnitude of the induced electromotive force (emf) in the wire, we can use the formula for motional emf, which is given by:

\[ \text{emf} = B \cdot L \cdot V \]

where:
- \( B \) is the magnetic field strength,
- \( L \) is the length of the wire (or aerial),
- \( V \) is the velocity of the boat.

### Step 1: Identify the given values
- Magnetic field strength, \( B = 5.0 \times 10^{-5} \, \text{N/A/m} \)
- Length of the aerial, \( L = 2 \, \text{m} \)
- Speed of the boat, \( V = 1.5 \, \text{m/s} \)

### Step 2: Substitute the values into the formula
Now, substituting the values into the formula for induced emf:

\[ \text{emf} = (5.0 \times 10^{-5} \, \text{N/A/m}) \cdot (2 \, \text{m}) \cdot (1.5 \, \text{m/s}) \]

### Step 3: Calculate the product
Calculating the product step by step:

1. Calculate \( 5.0 \times 10^{-5} \cdot 2 \):
\[ 5.0 \times 10^{-5} \cdot 2 = 1.0 \times 10^{-4} \]

2. Now multiply this result by the velocity \( 1.5 \):
\[ 1.0 \times 10^{-4} \cdot 1.5 = 1.5 \times 10^{-4} \]

### Step 4: Convert to millivolts
To express the induced emf in millivolts, we convert \( 1.5 \times 10^{-4} \) volts:
\[ 1.5 \times 10^{-4} \, \text{V} = 0.15 \, \text{mV} \]

### Final Answer
Thus, the magnitude of the induced emf in the wire is:
\[ \text{emf} = 0.15 \, \text{mV} \]