A horizontal straight wire 5 m long extending from east to west falling freely at right angle to horizontal component of earth’s magnetic field `0.60 times 10^(–4) Wbm^(–2)`. The instantaneous value of emf induced in the wire when its velocity is `10 ms^(–1)` is _________ `times 10^(–3) V`.

Answer

To find the instantaneous value of emf induced in the wire, we can use the formula for motional emf, which is given by:

\[ \text{emf} = B \cdot L \cdot V \cdot \sin(\theta) \]

Where:
- \( B \) = magnetic field strength (in Wb/m²)
- \( L \) = length of the wire (in meters)
- \( V \) = velocity of the wire (in m/s)
- \( \theta \) = angle between the direction of motion and the magnetic field (in degrees)

### Step 1: Identify the given values
From the problem statement, we have:
- Length of the wire, \( L = 5 \, \text{m} \)
- Magnetic field, \( B = 0.60 \times 10^{-4} \, \text{Wb/m}^2 \)
- Velocity of the wire, \( V = 10 \, \text{m/s} \)
- Since the wire is falling freely at right angles to the horizontal component of the magnetic field, \( \theta = 90^\circ \).

### Step 2: Calculate \(\sin(\theta)\)
Since \( \theta = 90^\circ \):
\[ \sin(90^\circ) = 1 \]

### Step 3: Substitute the values into the emf formula
Now, substituting the values into the emf formula:
\[ \text{emf} = B \cdot L \cdot V \cdot \sin(90^\circ) \]
\[ \text{emf} = (0.60 \times 10^{-4}) \cdot 5 \cdot 10 \cdot 1 \]

### Step 4: Perform the multiplication
Calculating the above expression:
\[ \text{emf} = 0.60 \times 10^{-4} \times 5 \times 10 \]
\[ = 0.60 \times 5 \times 10^{-4} \times 10 \]
\[ = 3.0 \times 10^{-4} \, \text{V} \]

### Step 5: Express the result in the required format
The problem asks for the answer in the form of \( x \times 10^{-3} \, \text{V} \). Therefore, we can write:
\[ 3.0 \times 10^{-4} \, \text{V} = 0.30 \times 10^{-3} \, \text{V} \]

Thus, the value of \( x \) is \( 3 \).

### Final Answer
The instantaneous value of emf induced in the wire is \( 3 \times 10^{-3} \, \text{V} \).

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