A conductor is moving with a uniform velocity of `10 m//s` at right angles to magenetic field of inducation `0.4 xx 10^(-4) Wb//m^(2)`. If the e.m.f. Induced in the conductor is `6 xx 10^(-5)V`, then the length of the conductor is

Options

A. 10 cm
B. 15 cm
C. 20 cm
D. 40 cm

Explanation

To find the length of the conductor, we can use the formula for the electromotive force (e.m.f.) induced in a moving conductor in a magnetic field:

\[ E = B \cdot L \cdot V \]

Where:
- \(E\) is the induced e.m.f. (in volts),
- \(B\) is the magnetic field induction (in Wb/m²),
- \(L\) is the length of the conductor (in meters),
- \(V\) is the velocity of the conductor (in m/s).

### Step-by-Step Solution:

1. **Identify the given values:**
- Induced e.m.f. \(E = 6 \times 10^{-5} \, V\)
- Magnetic field induction \(B = 0.4 \times 10^{-4} \, Wb/m^2\)
- Velocity \(V = 10 \, m/s\)

2. **Rearrange the formula to solve for \(L\):**
\[ L = \frac{E}{B \cdot V} \]

3. **Substitute the known values into the equation:**
\[ L = \frac{6 \times 10^{-5}}{(0.4 \times 10^{-4}) \cdot 10} \]

4. **Calculate the denominator:**
- First, calculate \(0.4 \times 10^{-4} = 4 \times 10^{-5}\).
- Now, \(B \cdot V = 4 \times 10^{-5} \cdot 10 = 4 \times 10^{-4}\).

5. **Now substitute back into the equation for \(L\):**
\[ L = \frac{6 \times 10^{-5}}{4 \times 10^{-4}} \]

6. **Perform the division:**
\[ L = \frac{6}{4} \times \frac{10^{-5}}{10^{-4}} = 1.5 \times 10^{-1} \, m \]

7. **Convert to centimeters:**
\[ L = 1.5 \times 10^{-1} \, m = 15 \, cm \]

### Final Answer:
The length of the conductor is \(15 \, cm\).

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