Question

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 $k$

Answer 1

Magnetic field at $P$ due to side $PQ$ and $RP$ are zero respectively

since point $P$ is on the sides $PQ$ and $RP$.

From the diagram in $\Delta PQS$ and $\Delta PSR$ sum of all the angles of a triangle are ${180}^{\circ }$;

$\angle QPS={37}^{\circ }$; $\angle SPR={53}^{\circ }$ since $\angle PRQ=\angle PSR={90}^{\circ }$
$PS=PQ\sin {53}^{\circ }=3x\times \frac{4}{5}=\frac{12x}{5}$

Magnetic field at P due to side QR:

$\overrightarrow{B}=\frac{{\mu }_{0}i}{4\pi PS}(\sin {53}^{\circ }+\sin {37}^{\circ })$

$=\frac{{\mu }_{0}i}{4\pi \times \frac{12x}{5}}(\frac{4}{5}+\frac{3}{5})=\frac{7{\mu }_{0}i}{48\pi x}$
$\therefore k=7$