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: Fix the direction first:
The field of a current element is perpendicular to the plane containing the element and the line to the point.
The element lies along the x-axis and the point lies along the y-axis from it. So the field must be along the z-axis.
This removes the options along $\hat{j}$, that is (A) and (D).

Step 2: Find the sign with the right-hand rule:
Use the right-hand rule: the current is along $+x$. Point the thumb along $+x$.
At a point on the $+y$ side, the fingers curl out of the page in the $+z$ direction.
This is the same as $\hat{i}\times\hat{j} = \hat{k}$. So the field is along $+\hat{k}$, and (C) is wrong.

Step 3: Find the size:
The Biot-Savart expression for the size, with $\theta = 90^{\circ}$, is $dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl}{r^2}$.
\[ dB = \left(10^{-7}\right)\frac{(10)(10^{-2})}{1^2} = 10^{-8}\ \text{T} \]
This is 10 nT, since 1 nT is $10^{-9}$ T. So the field is 10 nT along $+z$.

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