A rectangular loop of sides 'a' and 'b' is placed in xy plane. A very long wire is also placed in xy plane such that side of length 'a' of the loop is parallel to the wire. The distance between the wire and the nearest edge of the loop is 'd'. The mutual inductance of this system is proportional to :

Options

A. `a`
B. `b`
C. `1/d`
D. current in wire

Answer

To solve the problem of finding the mutual inductance of a rectangular loop placed parallel to a long wire, we can follow these steps:

### Step 1: Understand the Configuration
We have a rectangular loop with sides 'a' and 'b' placed in the xy-plane. A long wire is also in the xy-plane, parallel to the side of length 'a' of the loop. The distance between the wire and the nearest edge of the loop is 'd'.

**Hint:** Visualize the setup to understand the relationship between the loop and the wire.

### Step 2: Define the Current in the Wire
Let the current flowing through the long wire be denoted as \( I_0 \).

**Hint:** Identify the current as a variable that will affect the magnetic field.

### Step 3: Calculate the Magnetic Field
The magnetic field \( B \) at a distance \( x \) from the long wire can be expressed using Ampère's law:
\[ B = \frac{\mu_0 I_0}{2 \pi x} \]
where \( \mu_0 \) is the permeability of free space.

**Hint:** Remember that the magnetic field around a long straight conductor decreases with distance.

### Step 4: Consider a Small Element of the Loop
Take a small element of the loop, \( dA \), at a distance \( x \) from the wire. The area of this small element can be expressed as:
\[ dA = a \, dx \]
where \( dx \) is an infinitesimal width of the loop.

**Hint:** Break down the loop into small segments to simplify the integration process.

### Step 5: Calculate the Magnetic Flux
The magnetic flux \( d\Phi \) through the small area \( dA \) is given by:
\[ d\Phi = B \cdot dA = \left( \frac{\mu_0 I_0}{2 \pi x} \right) \cdot (a \, dx) \]
Thus, the total flux \( \Phi \) through the loop can be found by integrating \( d\Phi \) from \( d \) to \( b + d \):
\[ \Phi = \int_{d}^{b+d} \frac{\mu_0 I_0 a}{2 \pi x} \, dx \]

**Hint:** Set up the integral correctly to find the total magnetic flux through the loop.

### Step 6: Evaluate the Integral
The integral can be evaluated as follows:
\[ \Phi = \frac{\mu_0 I_0 a}{2 \pi} \left[ \ln(x) \right]_{d}^{b+d} = \frac{\mu_0 I_0 a}{2 \pi} \left( \ln(b+d) - \ln(d) \right) = \frac{\mu_0 I_0 a}{2 \pi} \ln\left(\frac{b+d}{d}\right) \]

**Hint:** Use properties of logarithms to simplify the expression.

### Step 7: Relate Flux to Mutual Inductance
The mutual inductance \( M \) is defined by the relation:
\[ \Phi = M I_0 \]
Thus, we can express \( M \) as:
\[ M = \frac{\Phi}{I_0} = \frac{\mu_0 a}{2 \pi} \ln\left(\frac{b+d}{d}\right) \]

**Hint:** Recognize that mutual inductance is proportional to the area of the loop.

### Step 8: Identify Proportionality
From the expression for mutual inductance \( M \), we can see that it is directly proportional to the side 'a' of the rectangle:
\[ M \propto a \]

**Hint:** Focus on the terms in the final expression to determine which variable affects mutual inductance.

### Conclusion
The mutual inductance of the system is proportional to the length 'a' of the rectangular loop.

**Final Answer:** Mutual inductance \( M \) is proportional to \( a \).