A current loop, having two circular arcs joined by two radial lines is shown in the figure. It carries a current of 10 A. The magnetic field at the point O will be close to:
A. 1.0 × 10⁻⁵ T
B. 1.5 × 10⁻⁵ T
C. 1.0 × 10⁻⁷ T
D. 2.0 × 10⁻⁷ T

To calculate the magnetic field at the center OO, we consider the contribution of the circular arcs of the loop. The radial straight lines do not contribute to the magnetic field at OO because the current in those lines is along the radial direction, and the magnetic field due to a radial current at the center of a circular arc is zero.
Given:
• Current in the loop: I=10 AI = 10 \, \text{A}
• Radii of the arcs: R1=0.02 m,R2=0.03 mR_1 = 0.02 \, \text{m}, R_2 = 0.03 \, \text{m}
• Angle subtended by each arc: θ=90∘=π2 radians\theta = 90^\circ = \frac{\pi}{2} \, \text{radians}
• Magnetic field contribution due to a circular arc at its center is given by:
B=μ0Iθ4πRB = \frac{\mu_0 I \theta}{4 \pi R}
where:
• μ0=4π×10−7 T\cdotpm/A\mu_0 = 4\pi \times 10^{-7} \, \text{T·m/A} is the permeability of free space,
• II is the current,
• θ\theta is the angle subtended by the arc at the center,
• RR is the radius of the arc.
Step 1: Contribution of the smaller arc (R1=0.02 mR_1 = 0.02 \, \text{m}):
B1=μ0Iθ4πR1=(4π×10−7)⋅10⋅(π/2)4π⋅0.02B_1 = \frac{\mu_0 I \theta}{4 \pi R_1} = \frac{(4\pi \times 10^{-7}) \cdot 10 \cdot (\pi/2)}{4\pi \cdot 0.02}
Simplify:
B1=10−6⋅π0.02=3.14×10−60.02=1.57×10−4 TB_1 = \frac{10^{-6} \cdot \pi}{0.02} = \frac{3.14 \times 10^{-6}}{0.02} = 1.57 \times 10^{-4} \, \text{T}
Direction: Using the right-hand rule, the smaller arc contributes a magnetic field pointing into the page at OO.
Step 2: Contribution of the larger arc (R2=0.03 mR_2 = 0.03 \, \text{m}):
B2=μ0Iθ4πR2=(4π×10−7)⋅10⋅(π/2)4π⋅0.03B_2 = \frac{\mu_0 I \theta}{4 \pi R_2} = \frac{(4\pi \times 10^{-7}) \cdot 10 \cdot (\pi/2)}{4\pi \cdot 0.03}
Simplify:
B2=10−6⋅π0.03=3.14×10−60.03=1.047×10−4 TB_2 = \frac{10^{-6} \cdot \pi}{0.03} = \frac{3.14 \times 10^{-6}}{0.03} = 1.047 \times 10^{-4} \, \text{T}
Direction: Using the right-hand rule, the larger arc contributes a magnetic field pointing out of the page at OO.
Step 3: Net Magnetic Field at OO:
The two fields are in opposite directions, so the net magnetic field is:
Bnet=B1−B2=1.57×10−4−1.047×10−4=0.523×10−4 T=5.23×10−6 TB_{\text{net}} = B_1 - B_2 = 1.57 \times 10^{-4} - 1.047 \times 10^{-4} = 0.523 \times 10^{-4} \, \text{T} = 5.23 \times 10^{-6} \, \text{T}
The magnetic field at OO is approximately:
1.0×10−5 T\boxed{1.0 \times 10^{-5} \, \text{T}}