The horizontal component of the earth's magnetic field at a place is `3xx10^(-4)T` and the dip is `tan^(-1)((4)/(3))`. A metal rod of length `0.25m` placed in the north -south position and is moved at a constant speed of `10cm//s` towards the east. The emf induced in the rod will be

Options

A. `1muV`
B. `5muV`
C. `7muV`
D. `10muV`

Explanation

To solve the problem, we need to calculate the induced electromotive force (emf) in a metal rod that is moved in a magnetic field. Here are the steps to find the solution:

### Step 1: Identify Given Values
- Horizontal component of Earth's magnetic field, \( B_h = 3 \times 10^{-4} \, \text{T} \)
- Length of the rod, \( L = 0.25 \, \text{m} \)
- Speed of the rod, \( v = 10 \, \text{cm/s} = 0.1 \, \text{m/s} \)
- Dip angle, \( \theta = \tan^{-1}\left(\frac{4}{3}\right) \)

### Step 2: Calculate the Vertical Component of the Magnetic Field
Using the dip angle, we can find the vertical component of the magnetic field \( B_v \):
\[ B_v = B_h \tan(\theta) \]
First, we need to calculate \( \tan(\theta) \):
\[ \tan(\theta) = \frac{4}{3} \]
Now substituting \( B_h \):
\[ B_v = B_h \cdot \tan(\theta) = 3 \times 10^{-4} \cdot \frac{4}{3} = 4 \times 10^{-4} \, \text{T} \]

### Step 3: Calculate the Induced EMF
The induced emf \( E \) in the rod can be calculated using the formula:
\[ E = B_v \cdot L \cdot v \]
Substituting the values we have:
\[ E = (4 \times 10^{-4}) \cdot (0.25) \cdot (0.1) \]
Calculating this gives:
\[ E = 4 \times 10^{-4} \cdot 0.25 \cdot 0.1 = 1 \times 10^{-6} \, \text{V} \]

### Step 4: Convert to Microvolts
Since \( 1 \, \text{V} = 10^6 \, \mu\text{V} \):
\[ E = 1 \, \mu\text{V} \]

### Final Answer
The induced emf in the rod is \( 1 \, \mu\text{V} \).

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