A rectangular loop of length `'l'` and breadth `'b'` is placed at a distance of `x` from an infinitely long wire carrying current `'i'` such that the direction of current is parallel to breadth. If the loop moves away from the current wire in a direction perpendicular to it with a velocity `'v'`, the magnitude of the e.m.f. in the loop is: (`mu_(0)=` permeability of free space)

Options

A. `(mu_(0)iv)/(2pix)((l+b)/(b))`
B. `(mu_(0)i^(2)v)/(4pi^(2)x)log((b)/(l))`
C. `(mu_(0)ilbv)/(2pix(l+x))`
D. `(mu_(0)ilbv)/(2pi)log((x+l)/(x))`

Explanation

To solve the problem, we need to determine the induced electromotive force (e.m.f.) in a rectangular loop that is moving away from an infinitely long wire carrying current. The steps to derive the expression for the induced e.m.f. are as follows:

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have a rectangular loop of length `l` and breadth `b`.
- The loop is placed at a distance `x` from an infinitely long wire carrying current `i`.
- The loop moves away from the wire with a velocity `v` in a direction perpendicular to the wire.

2. **Identify the Magnetic Field**:
- The magnetic field `B` around a long straight wire carrying current `i` at a distance `r` is given by:
\[ B = \frac{\mu_0 i}{2 \pi r} \]
- For the sides of the loop, we need to consider the magnetic field at two different distances:
- At distance `x` (for side AD)
- At distance `x + l` (for side BC)

3. **Calculate the Magnetic Fields**:
- The magnetic field at distance `x` (for side AD) is:
\[ B_1 = \frac{\mu_0 i}{2 \pi x} \]
- The magnetic field at distance `x + l` (for side BC) is:
\[ B_2 = \frac{\mu_0 i}{2 \pi (x + l)} \]

4. **Determine the Induced e.m.f.**:
- The induced e.m.f. (ε) in the loop can be calculated using the formula:
\[ \epsilon = b \cdot (B_1 - B_2) \cdot v \]
- Substituting the values of `B_1` and `B_2`:
\[ \epsilon = b \cdot \left(\frac{\mu_0 i}{2 \pi x} - \frac{\mu_0 i}{2 \pi (x + l)}\right) \cdot v \]

5. **Simplify the Expression**:
- Factor out common terms:
\[ \epsilon = \frac{\mu_0 i b v}{2 \pi} \left(\frac{1}{x} - \frac{1}{x + l}\right) \]
- Simplifying the expression inside the parentheses:
\[ \frac{1}{x} - \frac{1}{x + l} = \frac{(x + l) - x}{x(x + l)} = \frac{l}{x(x + l)} \]
- Therefore, the induced e.m.f. becomes:
\[ \epsilon = \frac{\mu_0 i b v l}{2 \pi x (x + l)} \]

### Final Result:
The magnitude of the induced e.m.f. in the loop is:
\[ \epsilon = \frac{\mu_0 i b v l}{2 \pi x (x + l)} \]