Question

A straight wire carrying a current of 12 A is bent into a semi-circular arc of radius 2 cm as shown in the figure. Then the magnetic field due to the straight segments at the centre of the arc is

• A. 12 T
• B. 6 T
• C. 24 T
• D. 0

Hint:

For magnetic fields from straight wires, use the Biot-Savart law and consider symmetry: fields from symmetric segments may cancel at the center of a semi-circular configuration.

Answer 1

The Correct Option is D

Step 1: Identify the straight segments and their contribution.
The wire is bent into a semi-circular arc of radius 2 cm, with straight segments extending infinitely on either side of the arc. The figure indicates a semi-circle centered at point \( O \), with the straight segments lying along a line (the diameter of the circle). We need the magnetic field at \( O \) due to the straight segments (not the arc). For a straight wire carrying current \( I \), the magnetic field at a perpendicular distance \( d \) is given by the Biot-Savart law or Ampere’s law: \[ B = \frac{\mu_0 I}{2 \pi d}. \]
Step 2: Analyze the geometry and calculate.
The straight segments are along the diameter of the semi-circle, and the center \( O \) is at a distance equal to the radius (2 cm = 0.02 m) from the straight wire. However, the key insight is the positioning: the straight segments are semi-infinite (extending from the ends of the arc to infinity). For a semi-infinite wire, the magnetic field at a point perpendicular to the wire at distance \( d \) is: \[ B = \frac{\mu_0 I}{4 \pi d}. \] Here, \( I = 12 \) A, \( d = 0.02 \) m. There are two straight segments (one from each end of the arc), both contributing to the field at \( O \). Compute for one segment: \[ B_1 = \frac{\mu_0 I}{4 \pi d} = \frac{(4 \pi \times 10^{-7}) \times 12}{4 \pi \times 0.02} = \frac{12 \times 10^{-7}}{0.02} = 6 \times 10^{-5} \, \text{T}. \] The other segment produces a field of the same magnitude but in the opposite direction (using the right-hand rule, the currents in the two straight segments produce fields at \( O \) that cancel out). Thus, the net field due to the straight segments is: \[ B_{\text{net}} = B_1 - B_1 = 0. \] Final Answer: The magnetic field due to the straight segments is \( \boxed{0} \).