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:
\includegraphics[width=0.5\linewidth]{3a.png}

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

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: - \( \mu_0 = 4\pi \times 10^{-7} \, }^2 \) is the magnetic constant. - \( I = 5.0 \, A} \) is the current. - \( dl = 0.02 \, m} \, \hat{j} \) 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): \[ r = \sqrt{3^2 + 4^2} = 5 \, m} \] \[ \hat{r} = \left(\frac{3}{5}, \frac{4}{5}, 0\right) \] 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: \[ dB = \frac{4\pi \times 10^{-7}}{4\pi} \cdot \frac{5 \cdot 0.012}{25} = \frac{10^{-7} \cdot 0.06}{25} = 2.4 \times 10^{-9} \, T} = 0.24 \, nT} \] Considering the direction, the field is \( -0.24 \, nT} \, \hat{k} \), matching option (C).