Question

A straight conductor lies along $x$-axis. It carries a current of 10 A along $+x$ direction. The magnetic field $\vec{B}$ due to 1 cm segment of this conductor, centred at the origin, at a point $(0, 1$ m$, 0)$ is:

• A. $(1\,\text{nT})\,\hat{j}$
• B. $(10\,\text{nT})\,\hat{k}$
• C. $-(10\,\text{nT})\,\hat{k}$
• D. $-(1\,\text{nT})\,\hat{j}$

Hint:

Use dB = (mu0/4pi) I dl x r-hat / r^2. The direction is i-hat cross j-hat, and the size is 1e-7 x 10 x 0.01.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The Biot-Savart law gives the magnetic field of a small current element.
A 1 cm segment is very short compared with the 1 m distance, so we can treat it as a single element \(d\vec{l}\).

Step 2: Key Formula or Approach:
\[ d\vec{B} = \frac{\mu_0}{4\pi}\,\frac{I\,d\vec{l}\times\hat{r}}{r^2} \]
Here \(I = 10\) A, \(dl = 1\ \text{cm} = 0.01\) m and \(r = 1\) m.
The element points along \(+\hat{i}\). The point is at \((0, 1, 0)\), so the unit vector from the element to the point is \(\hat{r} = +\hat{j}\).

Step 3: Detailed Explanation:
Magnitude:
\[ dB = \frac{10^{-7} \times 10 \times 0.01}{(1)^2} = 10^{-8}\ \text{T} = 10\ \text{nT} \]
Direction:
\[ \hat{i}\times\hat{j} = +\hat{k} \]
So the field is along \(+\hat{k}\), with size 10 nT.
\[ d\vec{B} = (10\ \text{nT})\,\hat{k} \]

Step 4: Check the options:
Option (A) is along \(\hat{j}\). A field from a current element is always perpendicular to both \(d\vec{l}\) and \(\hat{r}\). Here \(\hat{r}\) is already \(\hat{j}\), so this is impossible.
Option (C) has the right size but points along \(-\hat{k}\), which is the opposite direction.
Option (D) is along \(-\hat{j}\) and is also impossible.
Option (B) agrees with our calculation.

Final Answer:
The field at the point is \((10\ \text{nT})\,\hat{k}\), which is option (B). \[ \boxed{(10\ \text{nT})\,\hat{k}} \]