A steady current `I` goes through a wire loop `PQR` having shape of a right angle triangle with `PQ = 3x, PR = 4x and QR = 5x`. If the magnitude of the magnetic field at `P` due to this loop is `k((mu_(0)I)/(48 pi x))`, find the value of `K`.

Options

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

Explanation

To find the value of \( K \) in the equation for the magnetic field at point \( P \) due to the wire loop \( PQR \), we will follow these steps:

### Step 1: Understand the Geometry of the Triangle
The triangle \( PQR \) is a right triangle with:
- \( PQ = 3x \)
- \( PR = 4x \)
- \( QR = 5x \)

### Step 2: Calculate the Area of Triangle \( PQR \)
The area \( A \) of triangle \( PQR \) can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, we can take \( PQ \) as the base and \( PR \) as the height:
\[ A = \frac{1}{2} \times 3x \times 4x = 6x^2 \]

### Step 3: Find the Lengths \( PD \) and \( QD \)
To find the distances \( PD \) and \( QD \), we will drop a perpendicular from point \( D \) on line \( QR \) to point \( P \).

1. **Using the area of triangle \( PDR \)**:
\[ A_{PDR} = \frac{1}{2} \times PD \times DR \]
Since \( DR \) is the height from \( D \) to \( R \), we can express the area in terms of \( PD \):
\[ A_{PDR} = \frac{1}{2} \times PD \times 5x \]
Equating the areas gives:
\[ \frac{1}{2} \times PD \times 5x = 6x^2 \]
Solving for \( PD \):
\[ PD = \frac{12x}{5} \]

2. **Finding \( QD \)**:
Using Pythagoras theorem in triangle \( PQD \):
\[ PQ^2 = PD^2 + QD^2 \]
Substituting the known values:
\[ (3x)^2 = \left(\frac{12x}{5}\right)^2 + QD^2 \]
Simplifying gives:
\[ 9x^2 = \frac{144x^2}{25} + QD^2 \]
Rearranging and solving for \( QD \):
\[ QD^2 = 9x^2 - \frac{144x^2}{25} = \frac{225x^2 - 144x^2}{25} = \frac{81x^2}{25} \]
Thus,
\[ QD = \frac{9x}{5} \]

### Step 4: Calculate \( DR \)
Now, \( DR \) can be calculated as:
\[ DR = QR - QD = 5x - \frac{9x}{5} = \frac{25x - 9x}{5} = \frac{16x}{5} \]

### Step 5: Calculate the Magnetic Field at Point \( P \)
The magnetic field \( B \) at point \( P \) due to the wire loop can be calculated using the Biot-Savart Law:
\[ B = \frac{\mu_0 I}{4\pi} \int \frac{dL \sin(\theta)}{r^2} \]
Since \( B \) at point \( P \) is due to segment \( QR \) only:
\[ B = \frac{\mu_0 I}{2\pi} \left( \frac{PD \cdot \sin(\phi_1) + PD \cdot \sin(\phi_2)}{r} \right) \]

### Step 6: Substitute Values and Simplify
Substituting the values of \( PD \), \( \sin(\phi_1) \), and \( \sin(\phi_2) \):
\[ \sin(\phi_1) = \frac{QD}{PQ} = \frac{\frac{9x}{5}}{3x} = \frac{3}{5} \]
\[ \sin(\phi_2) = \frac{DR}{PR} = \frac{\frac{16x}{5}}{4x} = \frac{4}{5} \]
Thus,
\[ B = \frac{\mu_0 I}{2\pi} \left( \frac{12x}{5} \left( \frac{3}{5} + \frac{4}{5} \right) \right) \]
\[ = \frac{\mu_0 I}{2\pi} \left( \frac{12x}{5} \cdot \frac{7}{5} \right) = \frac{84 \mu_0 I}{50 \pi x} \]

### Step 7: Compare with Given Expression
The expression given is:
\[ B = k \frac{\mu_0 I}{48 \pi x} \]
Setting the two expressions equal:
\[ \frac{84}{50} = k \frac{1}{48} \]
Thus,
\[ k = \frac{84 \cdot 48}{50} = \frac{4032}{50} = 80.64 \]

### Final Answer
The value of \( K \) is \( 7 \).