Question

A loop ABCDA, carrying current I = 12 A, is placed in a plane, consists of two semi-circular segments of radius R₁ = 6π m and R₂ = 4π m. The magnitude of the resultant magnetic field at center O is k×$10^{-7}$ T. The value of k is ______. (Given μ₀ = 4π × $10^{-7}$ Tm A⁻¹)

Answer 1

1

Answer 2

For a semicircular arc carrying a current $I$, the magnetic field at the center is given by:

$$B = \frac{\mu_0 I}{4 R}$$

Step 1: Calculate the magnetic fields from each semicircular arc

• Outer Semicircle (radius $R_1 = 6\pi$)

$$B_1 = \frac{\mu_0 I}{4R_1} = \frac{4\pi \times 10^{-7} \times 12}{4 \times 6\pi} = \frac{48\pi \times 10^{-7}}{24\pi} = 2 \times 10^{-7}\ \text{T}$$

• Inner Semicircle (radius $R_2 = 4\pi$)

$$B_2 = \frac{\mu_0 I}{4R_2} = \frac{4\pi \times 10^{-7} \times 12}{4 \times 4\pi} = \frac{48\pi \times 10^{-7}}{16\pi} = 3 \times 10^{-7}\ \text{T}$$

Step 2: Determine the net magnetic field at $O$

The directions of the magnetic fields produced by the two arcs are opposite (due to the loop’s geometry). Thus, the net magnetic field is the difference in their magnitudes:

$$B_{\text{net}} = B_2 - B_1 = 3 \times 10^{-7} - 2 \times 10^{-7} = 1 \times 10^{-7}\ \text{T}$$

This gives $k \times 10^{-7}\ \text{T} = 1 \times 10^{-7}\ \text{T}$, so $k = 1$.

Core Explanation:

1. Use $B = \frac{\mu_0 I}{4R}$ for each semicircular arc.
2. Compute $B_1$ for $R_1 = 6\pi$ and $B_2$ for $R_2 = 4\pi$.
3. Since the fields are in opposite directions, net field $B_{\text{net}} = B_2 - B_1$.
4. Evaluate to get $k = 1$.