QJEEAdv09Ph5f5e: JEE Advanced 2009 Physics - Class 12 JEEAdv09Ph5f5e Solution: A steady current I goes through a wire loop PQR having shape of …

The segments PQ and PR, in the triangle shown here, cannot produce $\overrightarrow B$ at point P since point P lies on them. Here, only QR creates $\overrightarrow B$ at point P.

Applying ${B_P} = {{{\mu _0}i} \over {4\pi R}}(\cos {\phi _1} + \cos {\phi _2})$ , where ${\phi _1} = 53^\circ$ and ${\phi _2} = 37^\circ$ , we get the magnitude of the magnetic field as follows:

${B_P} = {{{\mu _0}I} \over {4\pi (4\pi \sin 37^\circ )}}(\cos 53^\circ + \cos 37^\circ )$

$= {{{\mu _0}I} \over {16\pi x(3/5)}}\left( {{{3x} \over {5x}} + {{4x} \over {5x}}} \right)$

$= {{5{\mu _0}I} \over {48\pi x}}\left( {{7 \over 5}} \right) = {{7{\mu _0}I} \over {48\pi x}} = k\left( {{{{\mu _0}I} \over {48\pi x}}} \right)$

where $k = 7$ .