A long straight wire carrying current I and a square conducting wire loop of side l , at a distance 'a' from current wire as shown in the figure. Both the current wire and loop are in the plane of paper. Find the magnetic flux of current wire, passing through the loop.

Options

A. `(mu_(0))/(2pi).Illog_(e).(a+I)/(a)`
B. `(mu_(0))/(4pi).Illog_(e).(a+I)/(a)`
C. `(mu_(0))/(2pi).Il`
D. `(mu_(0))/(4pi).I`

Explanation

To find the magnetic flux of a long straight wire carrying current \( I \) passing through a square conducting wire loop of side \( l \) at a distance \( a \) from the current wire, we can follow these steps:

### Step 1: Determine the Magnetic Field due to the Long Straight Wire
The magnetic field \( B \) at a distance \( r \) from a long straight wire carrying current \( I \) is given by the formula:
\[ B = \frac{\mu_0 I}{2 \pi r} \]
where \( \mu_0 \) is the permeability of free space.

### Step 2: Set Up the Geometry
The square loop is positioned such that its closest side is at a distance \( a \) from the wire. The distance from the wire to any point on the loop varies from \( a \) to \( a + l \) as we move from the closest side to the farthest side of the loop.

### Step 3: Calculate the Magnetic Field at Different Points on the Loop
For a small strip of the loop at a distance \( x \) from the wire (where \( x \) ranges from \( a \) to \( a + l \)), the magnetic field \( B \) can be expressed as:
\[ B(x) = \frac{\mu_0 I}{2 \pi x} \]

### Step 4: Calculate the Differential Area Element
The differential area \( dA \) for a small strip of width \( dx \) and height \( l \) is:
\[ dA = l \, dx \]

### Step 5: Calculate the Differential Magnetic Flux
The differential magnetic flux \( d\Phi \) through the small strip is given by:
\[ d\Phi = B(x) \cdot dA = \left(\frac{\mu_0 I}{2 \pi x}\right) l \, dx \]

### Step 6: Integrate to Find Total Magnetic Flux
To find the total magnetic flux \( \Phi \) through the entire loop, we integrate \( d\Phi \) from \( x = a \) to \( x = a + l \):
\[ \Phi = \int_{a}^{a+l} \frac{\mu_0 I l}{2 \pi x} \, dx \]

### Step 7: Perform the Integration
The integral of \( \frac{1}{x} \) is \( \ln(x) \):
\[ \Phi = \frac{\mu_0 I l}{2 \pi} \left[ \ln(x) \right]_{a}^{a+l} \]
\[ \Phi = \frac{\mu_0 I l}{2 \pi} \left( \ln(a + l) - \ln(a) \right) \]
Using the property of logarithms \( \ln(m) - \ln(n) = \ln\left(\frac{m}{n}\right) \):
\[ \Phi = \frac{\mu_0 I l}{2 \pi} \ln\left(\frac{a + l}{a}\right) \]

### Final Result
The magnetic flux \( \Phi \) through the square loop is:
\[ \Phi = \frac{\mu_0 I l}{2 \pi} \ln\left(\frac{a + l}{a}\right) \]