Animated Solution for Physics - Magnetic Effects of Current: 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 $k\left(\frac{{\mu }_{0}I}{48\pi x}\right)$, find the value of $k$.

The beauty of physics often lies in how elegantly it intertwines with pure geometry. This problem is a perfect example of that synergy. We are given a triangular loop $PQR$ carrying a steady current $I$, and we need to determine the magnetic field at one of its vertices, $P$.

The first thing to notice is the lengths of the sides of the triangle: $PQ=3x$, $PR=4x$, and $QR=5x$. Any student of geometry will immediately recognize this as a classic 3-4-5 Pythagorean triplet. This tells us that the triangle is a right-angled triangle, with the right angle located at vertex $P$ (since it is opposite the longest side, $QR$).

Now, let's apply the Biot-Savart law. The magnetic field $d\mathbf{B}$ produced by a small current element $d\mathbf{l}$ at a position $\mathbf{r}$ is proportional to the cross product $d\mathbf{l}\times \mathbf{r}$. For any point lying directly on the axis of a straight wire, the angle between $d\mathbf{l}$ and $\mathbf{r}$ is either $0^{\circ }$ or $180^{\circ }$. Since the sine of both these angles is zero, the cross product vanishes.

Because point $P$ lies exactly on the lines extending from wires $PQ$ and $PR$, these two segments contribute absolutely nothing to the magnetic field at $P$.

This simplifies our problem immensely. The entire magnetic field at point $P$ is generated solely by the hypotenuse, wire $QR$. To find the magnetic field produced by a finite straight wire at a point, we use the standard formula:

Here, $r$ is the perpendicular distance from the point to the wire, and ${\theta }_1$ and ${\theta }_2$ are the angles subtended by the ends of the wire at the foot of the perpendicular.

To use our master equation, we need to find $r$, ${\theta }_1$, and ${\theta }_2$. Let's drop a perpendicular from $P$ to the hypotenuse $QR$ and call its length $r$.

First, let's find the angles of the main triangle. Using basic trigonometry:

Now, look at the smaller right-angled triangle formed by $P$, $Q$, and the foot of the perpendicular. The angle at $P$ inside this smaller triangle is $90^{\circ }-53^{\circ }=37^{\circ }$. Therefore, the perpendicular distance $r$ can be found using the cosine function:

The angles subtended by the ends of the wire $QR$ at the foot of the perpendicular are simply the angles inside the smaller right triangles at $P$, which are $37^{\circ }$ and $53^{\circ }$.

Now we have everything we need. Let's substitute $r=\frac{12x}{5}$, ${\theta }_1=37^{\circ }$, and ${\theta }_2=53^{\circ }$ into our magnetic field formula:

The $5$ in the numerator and denominator cancel out beautifully:

The problem states that the magnetic field is $k\left(\frac{{\mu }_{0}I}{48\pi x}\right)$. By comparing our result with the given expression, it is clear that the integer value we are looking for is $k=7$.