Magnetic Field at Right Angle Vertex of Triangular Wire Loop

To find the magnitude of the magnetic field at point $\text{P}$ due to the triangular wire loop $\text{PQR}$, we consider the contribution from each of the three straight segments: $\text{PQ}$, $\text{PR}$, and $\text{QR}$.

1. Contribution from segments $\text{PQ}$ and $\text{PR}$:
   According to the Biot-Savart Law, the magnetic field produced by a straight current-carrying wire at any point lying along the line of the wire is zero. Since point $\text{P}$ lies on the line passing through segment $\text{PQ}$ and also on the line passing through segment $\text{PR}$, the magnetic fields at $\text{P}$ due to these segments are:
   $B_{\text{PQ}} = 0$
   $B_{\text{PR}} = 0$

2. Contribution from segment $\text{QR}$:
   The triangle $\text{PQR}$ is a right-angled triangle at $\text{P}$ because:
   $\text{PQ}^2 + \text{PR}^2 = (3x)^2 + (4x)^2 = 25x^2 = (5x)^2 = \text{QR}^2$

Let $d$ be the perpendicular distance from vertex $\text{P}$ to the hypotenuse $\text{QR}$.
The area of $\triangle \text{PQR}$ can be written in two ways:
$\text{Area} = \frac{1}{2} \times \text{PQ} \times \text{PR} = \frac{1}{2} \times (3x) \times (4x) = 6x^2$
$\text{Area} = \frac{1}{2} \times \text{QR} \times d = \frac{1}{2} \times (5x) \times d$

Equating the two expressions for area:
$\frac{5x}{2} d = 6x^2 \implies d = \frac{12x}{5}$

Let $\text{H}$ be the foot of the perpendicular from $\text{P}$ onto $\text{QR}$. Let $\theta_1$ and $\theta_2$ be the angles subtended by the segments $\text{HQ}$ and $\text{HR}$ at $\text{P}$, respectively:
$\theta_1 = \angle \text{HPQ} \quad \text{and} \quad \theta_2 = \angle \text{HPR}$

In the right-angled triangle $\text{PHQ}$:
$\sin \theta_1 = \cos(\angle \text{PQH}) = \frac{\text{PQ}}{\text{QR}} = \frac{3x}{5x} = \frac{3}{5}$

In the right-angled triangle $\text{PHR}$:
$\sin \theta_2 = \cos(\angle \text{PRH}) = \frac{\text{PR}}{\text{QR}} = \frac{4x}{5x} = \frac{4}{5}$

The magnetic field at $\text{P}$ due to the segment $\text{QR}$ is given by:
$B = \frac{\mu_0 I}{4\pi d} (\sin \theta_1 + \sin \theta_2)$

Substituting the values of $d$, $\sin \theta_1$, and $\sin \theta_2$:
$B = \frac{\mu_0 I}{4\pi \left(\frac{12x}{5}\right)} \left(\frac{3}{5} + \frac{4}{5}\right)$
$B = \frac{5\mu_0 I}{48\pi x} \left(\frac{7}{5}\right) = 7\left(\frac{\mu_0 I}{48\pi x}\right)$

Comparing this result with the given expression $B = k\left(\frac{\mu_0 I}{48\pi x}\right)$, we find:
$k = 7$