Question

A 2.0 cm segment of wire, carrying 5.0 A current in positive y-direction lies along the y-axis, as shown in the figure. The magnetic field at a point (3 m, 4 m, 0) due to this segment (part of a circuit) is:

 

• A. \((0.12 \, \text{nT}) \, \hat{j}\)
• B. \(-(0.10 \, \text{nT}) \, \hat{j}\)
• C. \(-(0.24 \, \text{nT}) \, \hat{k}\)
• D. \((0.24 \, \text{nT}) \, \hat{k}\)

Hint:

When dealing with the Biot-Savart Law, always remember to apply the right-hand rule to determine the direction of \( dl \times \hat{r} \) effectively, as it dictates the direction of the magnetic field produced by the current segment.

Answer 1

The Correct Option is C

Step 1: Application of the Biot-Savart Law.
The magnetic field due to a segment of current-carrying wire is determined by the Biot-Savart Law, which states: \[ dB = \frac{\mu_0}{4\pi} \frac{I \, dl \times \hat{r}}{r^2} \] where:

• \(\mu_0 = 4\pi \times 10^{-7}\ \mathrm{T\,m\,A^{-1}}\) is the magnetic constant.
• \( I = 5.0 \, \text{A} \) is the current.
• \( dl = 0.02 \, \text{m} \, \hat{j} \) is the length vector of the wire segment.
• \( \hat{r} \) is the unit vector from the segment to the observation point.

Step 2: Calculation of \( r \) and \( \hat{r} \).
From the wire segment to the point (3 m, 4 m, 0): \[ r = \sqrt{3^2 + 4^2} = 5 \, \text{m} \] \[ \hat{r} = \left(\frac{3}{5}, \frac{4}{5}, 0\right) \]

Step 3: Calculation of \( dl \times \hat{r} \) and \( dB \).
Using the right-hand rule for the cross product: \[ dl \times \hat{r} = 0.02 \, \hat{j} \times \left(\frac{3}{5}, \frac{4}{5}, 0\right) = 0.02 \left( 0, 0, -\frac{3}{5} \right) = \left( 0, 0, -0.012 \right) \, \hat{k} \] Thus, \( dB \) points in the negative \( \hat{k} \) direction. The magnitude of \( dB \) is: \[ dB = \frac{4\pi \times 10^{-7}}{4\pi} \cdot \frac{5 \cdot 0.012}{25} = \frac{10^{-7} \cdot 0.06}{25} = 2.4 \times 10^{-9} \, \text{T} = 0.24 \, \text{nT} \] Considering the direction, the field is: \[ -0.24 \, \text{nT} \, \hat{k} \] which matches option (C).

Answer 2

 

Step-by-Step Solution

Step 1: Biot–Savart law

dB = (μ₀/4π) · I · (dl × r) / r³

Step 2: Current element and position vector:

dl = ⟨0, dy, 0⟩ r  = ⟨3, 4−y, 0⟩

Step 3: Cross product:

dl × r = ⟨0,0, −3 dy⟩ = −3 dy k̂

Direction is along −k̂.

Step 4: Integral setup:

B = (μ₀ I / 4π)(−3 k̂) ∫₀^{0.02} dy / [9 + (4−y)²]^(3/2)

Step 5: Evaluate integral:

Result ≈ 1.6077×10⁻⁴ m⁻²

Step 6: Magnitude:

|B| = (1×10⁻⁷)(5)(3)(1.6077×10⁻⁴)    ≈ 2.41×10⁻¹⁰ T = 0.241 nT

Final Answer: B = −(0.24 nT) k̂ → Option (iii)

Key Point

• Direction from right-hand rule: −k̂
• Magnitude is tiny since wire is very short and point is far