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NEET PHYSICSMOVING CHARGES AND MAGNETISMMedium

Question

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

A

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

B

\frac{\mu_0 i}{2 R}

C

\frac{\mu_0 i}{4 R}

D

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

Step-by-Step Solution

  1. Magnetic Field of a Semicircle: The magnetic field at the center of a full circular loop carrying current ii is B=μ0i2RB = \frac{\mu_0 i}{2R}. For a semicircular arc (half a loop), the magnitude is half of this: Bsemi=12μ0i2R=μ0i4RB_{semi} = \frac{1}{2} \frac{\mu_0 i}{2R} = \frac{\mu_0 i}{4R}.
  2. Direction: The direction of the magnetic field follows the right-hand rule .
  • For the semicircle in the x−yx-y plane, the axis is the zz-axis, so the field B⃗1\vec{B}_1 is along the zz-direction (k^\hat{k}).
  • For the semicircle in the x−zx-z plane, the axis is the yy-axis, so the field B⃗2\vec{B}_2 is along the yy-direction (j^\hat{j}).
  1. Superposition: The two fields are perpendicular to each other. The resultant magnitude BnetB_{net} is the vector sum: Bnet=B12+B22=(μ0i4R)2+(μ0i4R)2B_{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} Bnet=μ0i4R2=μ0i22RB_{net} = \frac{\mu_0 i}{4R} \sqrt{2} = \frac{\mu_0 i}{2\sqrt{2} R} .

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from MOVING CHARGES AND MAGNETISM. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSMOVING CHARGES AND MAGNETISMcurrentconsistsidenticalsemicircularradius

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