Question:

A current-carrying wire loop lies in one plane and consists of a semicircle of radius 10.0 cm, a smaller semicircle with the same center, and two radial straight segments. The smaller semicircle is rotated out of the plane by an angle θ, until it is perpendicular to the plane. A graph shows the magnetic field at the center of curvature as a function of θ. The vertical scale is defined by Ba = 10.0 μT and Bb = 12.0 μT.

What is the radius of the smaller semicircle?

Answer with Detailed Explanation:

Step 1: Understanding Magnetic Field Contributions

Only the curved portions of the wire (semicircles) contribute to the magnetic field at the center.

• Magnetic field by a semicircular arc: B = μ₀I / 4R
• The larger semicircle lies in the plane and contributes fully: B₁ = μ₀I / 4R
• The smaller semicircle is rotated by angle θ, so only its vertical component contributes: B₂ = (μ₀I / 4r) × cos(θ)

The net magnetic field at the center becomes:
B(θ) = B₁ − B₂ × cos(θ)

Step 2: Use Given Values from the Graph

• At θ = 0°: B = Ba = B₁ − B₂
• At θ = 90°: B = Bb = B₁ (since cos(90°) = 0)

Step 3: Determine μ₀I

From Bb = μ₀I / 4R and R = 0.10 m:
μ₀I = 4 × Bb × R = 4 × 12.0 × 10⁻⁶ T × 0.10 m = 4.8 × 10⁻⁶ T·m

Step 4: Use ΔB to Find r

At θ = 0°:
Ba = Bb − μ₀I / 4r

Solving for r:
μ₀I / 4r = Bb − Ba = 12.0 μT − 10.0 μT = 2.0 μT = 2.0 × 10⁻⁶ T
4r = μ₀I / 2.0 × 10⁻⁶
r = (4.8 × 10⁻⁶) / (8.0 × 10⁻⁶) = 0.6 m = 6.0 cm

Conclusion:

The radius of the smaller semicircle is 6.0 cm.

Final Answer:

Radius of the smaller semicircle: 6.0 cm