Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A current loop consists of two identical semicircular parts each of radius $R$ , one lying in the $x-y$ plane, and the other in the $x-z$ plane. If the current in the loop is $i$ , what will be the resultant magnetic field due to the two semicircular parts at their common centre?

\frac{\mu_0 i}{2\sqrt{2} R}

\frac{\mu_0 i}{2 R}

\frac{\mu_0 i}{4 R}

\frac{\mu_0 i}{\sqrt{2} R}

Step-by-Step Solution

Magnetic Field of a Semicircle: The magnetic field at the center of a full circular loop carrying current $i$ is $B = \frac{\mu_0 i}{2R}$ . For a semicircular arc (half a loop), the magnitude is half of this: $B_{semi} = \frac{1}{2} \frac{\mu_0 i}{2R} = \frac{\mu_0 i}{4R}$ .
Direction: The direction of the magnetic field follows the right-hand rule .

For the semicircle in the $x-y$ plane, the axis is the $z$ -axis, so the field $\vec{B}_1$ is along the $z$ -direction ( $\hat{k}$ ).
For the semicircle in the $x-z$ plane, the axis is the $y$ -axis, so the field $\vec{B}_2$ is along the $y$ -direction ( $\hat{j}$ ).

Superposition: The two fields are perpendicular to each other. The resultant magnitude $B_{net}$ is the vector sum: $B_{net} = \sqrt{B_1^2 + B_2^2} = \sqrt{\left(\frac{\mu_0 i}{4R}\right)^2 + \left(\frac{\mu_0 i}{4R}\right)^2}$ $B_{net} = \frac{\mu_0 i}{4R} \sqrt{2} = \frac{\mu_0 i}{2\sqrt{2} R}$ .

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