A metal bar of length `1 m` falls from rest under the action of gravity remaining horizontal with its ends in east-west direction. The induced `e.m.f` in it at the instant when it fallen for `10 s` is `(B_(H) = 1.7 xx 10^(-5)T and g = 10m s^(-2))`

Options

A. `2.5m V`
B. `3.2 m V`
C. `1.7 m V`
D. `0.5 m V`

Explanation

To find the induced electromotive force (e.m.f) in a metal bar falling under the influence of gravity, we can follow these steps:

### Step 1: Determine the velocity of the bar after 10 seconds
The bar falls under the influence of gravity, which means it accelerates downwards with an acceleration of \( g = 10 \, \text{m/s}^2 \).

Using the equation of motion:
\[ v = u + gt \]
where:
- \( u = 0 \) (initial velocity),
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity),
- \( t = 10 \, \text{s} \) (time).

Substituting the values:
\[ v = 0 + (10 \, \text{m/s}^2)(10 \, \text{s}) = 100 \, \text{m/s} \]

### Step 2: Calculate the induced e.m.f
The induced e.m.f (\( \mathcal{E} \)) in a conductor moving through a magnetic field can be calculated using the formula:
\[ \mathcal{E} = B \cdot l \cdot v \]
where:
- \( B \) is the magnetic field strength,
- \( l \) is the length of the conductor,
- \( v \) is the velocity of the conductor.

Given:
- \( B = 1.7 \times 10^{-5} \, \text{T} \) (horizontal component of the Earth's magnetic field),
- \( l = 1 \, \text{m} \) (length of the metal bar),
- \( v = 100 \, \text{m/s} \) (velocity of the bar).

Substituting the values:
\[ \mathcal{E} = (1.7 \times 10^{-5} \, \text{T}) \cdot (1 \, \text{m}) \cdot (100 \, \text{m/s}) = 1.7 \times 10^{-3} \, \text{V} \]

### Step 3: Convert the e.m.f to millivolts
To express the e.m.f in millivolts:
\[ 1.7 \times 10^{-3} \, \text{V} = 1.7 \, \text{mV} \]

### Final Answer
The induced e.m.f in the metal bar after falling for 10 seconds is \( 1.7 \, \text{mV} \).

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