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A very long conducting wire is bent in a semi-circular shape from A to B as shown in the figure. The magnetic field at the point P for steady current configuration is given by:

A

\dfrac{\mu_0 i}{4R}\left[1-\dfrac{2}{\pi}\right] \text{ pointed into the page}

B

\dfrac{\mu_0 i}{4R} \text{ pointed into the page}

C

\dfrac{\mu_0 i}{4R} \text{ pointed away from the page}

D

\dfrac{\mu_0 i}{4R}\left[1-\dfrac{2}{\pi}\right] \text{ pointed away from the page}

Step-by-Step Solution

  1. Analyze the Geometry: The wire consists of three segments: two semi-infinite straight wires and one semi-circular arc. Based on the answer format (1−2/π1 - 2/\pi), the fields from the straight segments must oppose the field from the arc.
  2. Field due to Straight Segments: Consider the two semi-infinite straight wires extending to infinity. The magnetic field B1B_1 due to a semi-infinite wire at a perpendicular distance RR from one end is given by B1=μ0I4πRB_1 = \frac{\mu_0 I}{4\pi R}. Since there are two such wires carrying current in directions that produce magnetic fields in the same direction (opposing the arc), their combined field is: Bstraight=2×μ0I4πR=μ0I2πRB_{straight} = 2 \times \frac{\mu_0 I}{4\pi R} = \frac{\mu_0 I}{2\pi R}
  3. Field due to Semi-Circular Arc: The magnetic field BarcB_{arc} at the center of a semi-circular arc of radius RR 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}
  4. Net Magnetic Field: The net magnetic field is the difference between the arc field and the straight wire field (assuming the standard configuration where they oppose): Bnet=Barc−Bstraight=μ0I4R−μ0I2πRB_{net} = B_{arc} - B_{straight} = \frac{\mu_0 I}{4R} - \frac{\mu_0 I}{2\pi R} Factor out μ0I4R\frac{\mu_0 I}{4R}: Bnet=μ0I4R(1−2π)B_{net} = \frac{\mu_0 I}{4R} \left( 1 - \frac{2}{\pi} \right)
  5. Direction: The term 11 (from the arc) is greater than 2π≈0.63\frac{2}{\pi} \approx 0.63 (from the wires). Therefore, the net field direction is determined by the arc. If the arc current is counter-clockwise, the field points out of the page (away). The option 'pointed away from the page' corresponds to this standard case.
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