Step-by-step Derivations:

1. Analyze the Geometry of the Wire Loop: The wire loop PQR is in the shape of a right-angled triangle. The side lengths are given as PQ = 3x, PR = 4x, and QR = 5x. Let's verify if it's a right-angled triangle using the Pythagorean theorem: $(PQ)^2 + (PR)^2 = (3x)^2 + (4x)^2 = 9x^2 + 16x^2 = 25x^2$ $(QR)^2 = (5x)^2 = 25x^2$ Since $(PQ)^2 + (PR)^2 = (QR)^2$, the triangle is indeed a right-angled triangle with the right angle at vertex P.

2. Apply the Principle of Superposition: The total magnetic field at point P, denoted as $B_P$, is the vector sum of the magnetic fields produced by the three segments of the wire: PQ, PR, and QR. $B_P = B_{PQ} + B_{PR} + B_{QR}$

3. Calculate the Magnetic Field from Segments PQ and PR: The magnetic field due to a straight current-carrying wire at any point lying on the line of the wire (or its extension) is zero. This is because, according to the Biot-Savart law, $d\vec{B} = (\mu_0 I / 4\pi) (d\vec{l} \times \vec{r}) / r^3$, the vector element $d\vec{l}$ and the position vector $\vec{r}$ are collinear, making their cross product $d\vec{l} \times \vec{r}$ equal to zero.

   • For the segment PQ, point P lies on the axis of the wire. Therefore, the magnetic field at P due to PQ is zero: $B_{PQ} = 0$.
   • Similarly, for the segment PR, point P lies on the axis of the wire. Therefore, the magnetic field at P due to PR is zero: $B_{PR} = 0$.

4. Calculate the Magnetic Field from Segment QR: The total magnetic field at P is solely due to the segment QR: $B_P = B_{QR}$. The magnitude of the magnetic field at a point due to a finite straight wire is given by: $B = \frac{\mu_0 I}{4\pi d}(\cos\theta_1 + \cos\theta_2)$ where d is the perpendicular distance from the point to the wire, and $\theta_1$ and $\theta_2$ are the angles that the lines joining the point to the ends of the wire make with the wire itself.

   • Find the perpendicular distance d: Let d be the length of the altitude from P to the hypotenuse QR. We can find d by equating the area of the triangle PQR calculated in two ways: $Area = (1/2) \times \text{base} \times \text{height} = (1/2) \times PQ \times PR = (1/2)(3x)(4x) = 6x^2$ $Area = (1/2) \times QR \times d = (1/2)(5x)d$ Equating the two expressions for the area: $6x^2 = \frac{1}{2}(5x)d \implies d = \frac{12x^2}{5x} = \frac{12x}{5}$

   • Find the angles $\theta_1$ and $\theta_2$: In our case, $\theta_1 = \angle PQR$ and $\theta_2 = \angle PRQ$. From the right-angled triangle PQR: $\cos(\angle PQR) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{PQ}{QR} = \frac{3x}{5x} = \frac{3}{5}$ $\cos(\angle PRQ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{PR}{QR} = \frac{4x}{5x} = \frac{4}{5}$

   • Calculate $B_{QR}$: Substitute the values of d, $\cos\theta_1$, and $\cos\theta_2$ into the magnetic field formula: $B_{QR} = \frac{\mu_0 I}{4\pi (12x/5)} \left( \frac{3}{5} + \frac{4}{5} \right)$ $B_{QR} = \frac{5\mu_0 I}{48\pi x} \left( \frac{7}{5} \right)$ $B_{QR} = \frac{7\mu_0 I}{48\pi x}$

5. Determine the Value of k: The magnitude of the total magnetic field at P is $|B_P| = |B_{QR}| = \frac{7\mu_0 I}{48\pi x}$. The problem states that this magnitude is equal to $k\left( \frac{\mu_0 I}{48\pi x} \right)$. Comparing our result with the given expression: $k\left( \frac{\mu_0 I}{48\pi x} \right) = 7\left( \frac{\mu_0 I}{48\pi x} \right)$ Therefore, the value of k is 7.

Final Answer:

The value of k is 7.