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NEET PHYSICSMedium

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})
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