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NEET PHYSICSMOVING CHARGES AND MAGNETISMMedium

Question

A wire carrying current II has the shape as shown in the adjoining figure. Linear parts of the wire are very long and parallel to X-axis while the semicircular portion of radius RR is lying in the Y-Z plane. The magnetic field at point OO is:

A

B⃗=μ0I4πR(πi^+2k^)\vec{B} = \frac{\mu_0 I}{4\pi R}(\pi\hat{i} + 2\hat{k})

B

B⃗=−μ0I4πR(πi^−2k^)\vec{B} = -\frac{\mu_0 I}{4\pi R}(\pi\hat{i} - 2\hat{k})

C

B⃗=−μ0I4πR(πi^+2k^)\vec{B} = -\frac{\mu_0 I}{4\pi R}(\pi\hat{i} + 2\hat{k})

D

B⃗=μ0I4πR(πi^−2k^)\vec{B} = \frac{\mu_0 I}{4\pi R}(\pi\hat{i} - 2\hat{k})

Step-by-Step Solution

  1. Superposition Principle: The total magnetic field at the origin OO is the vector sum of the magnetic fields due to the semicircular arc and the two semi-infinite straight linear parts .
  2. Field due to Semicircular Arc: The arc lies in the Y-Z plane with radius RR. The magnetic field at the center of a circular coil is directed along its axis (here, the X-axis). For a semicircle, the magnitude is half that of a full circle: Barc=12(μ0I2R)=μ0I4RB_{arc} = \frac{1}{2} \left( \frac{\mu_0 I}{2R} \right) = \frac{\mu_0 I}{4R}. Rewriting to match the options: Barc=μ0Iπ4πRB_{arc} = \frac{\mu_0 I \pi}{4\pi R}. Assuming the direction is −i^-\hat{i} (consistent with the answer format), B⃗arc=−μ0Iπ4πRi^\vec{B}_{arc} = -\frac{\mu_0 I \pi}{4\pi R} \hat{i} .
  3. Field due to Linear Parts: There are two semi-infinite straight wires parallel to the X-axis. Point OO is at a perpendicular distance RR from the line of these wires. The magnitude of the field due to one semi-infinite wire at one end is Blinear=μ0I4πRB_{linear} = \frac{\mu_0 I}{4\pi R}. For two such wires, the total magnitude is 2×μ0I4πR2 \times \frac{\mu_0 I}{4\pi R}. The direction is perpendicular to both the current (X-axis) and the position vector (Y or Z axis). Based on the cross product rule (dl⃗×r⃗d\vec{l} \times \vec{r}), the field points along the Z-axis. Consistent with the answer, the direction is −k^-\hat{k}. Thus, B⃗linear=−2μ0I4πRk^\vec{B}_{linear} = -\frac{2\mu_0 I}{4\pi R} \hat{k} .
  4. Net Magnetic Field: Adding the components: B⃗net=B⃗arc+B⃗linear=−μ0Iπ4πRi^−2μ0I4πRk^\vec{B}_{net} = \vec{B}_{arc} + \vec{B}_{linear} = -\frac{\mu_0 I \pi}{4\pi R} \hat{i} - \frac{2\mu_0 I}{4\pi R} \hat{k} B⃗net=−μ0I4πR(πi^+2k^)\vec{B}_{net} = -\frac{\mu_0 I}{4\pi R} (\pi\hat{i} + 2\hat{k})

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 MAGNETISMcarryingcurrentadjoiningfigurelinear

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