Question

Two infinitely long straight conductors lie along the x-axis and y-axis in the plane of the page. If they carry current $I$ each along $+x$-axis and $+y$-axis respectively, the net magnetic field at a point with position vector $\vec{r} = (3\hat{i} + 4\hat{j})$ m will be:

• A. $\dfrac{\mu_0 I}{6\pi}\hat{k}$
• B. $-\dfrac{\mu_0 I}{12\pi}\hat{k}$
• C. $-\dfrac{\mu_0 I}{24\pi}\hat{k}$
• D. $\dfrac{\mu_0 I}{48\pi}\hat{k}$

Hint:

Find each wire's field using \(\mu_0I/2\pi d\) with d = 4 m and 3 m, give directions by the right-hand rule, and add. The result is \(-\mu_0I/24\pi\) along k.

Answer 1

The Correct Option is C

Step 1: Use superposition:
The field at the point is the sum of the fields of the two wires, each found as if the other were absent. For a long wire, $B=\mu_0I/(2\pi d)$.

Step 2: Directions with a simple picture:
Point your right thumb along the current and curl your fingers. For the wire on the x-axis, the fingers go out of the page above the wire. The point has $y=4$ m, which is above the x-axis, so this field is along $+\hat k$.
For the wire on the y-axis, the fingers go into the page on the right side of the wire. The point has $x=3$ m, which is on the right, so this field is along $-\hat k$.

Step 3: Compare sizes:
The distance to the x-axis wire is 4 m, and to the y-axis wire is 3 m. The nearer wire gives the stronger field. So the net field follows the y-axis wire and points along $-\hat k$.
This already rules out options (A) and (D), which are positive.

Step 4: Calculate:
\[ B_{net}=\frac{\mu_0I}{2\pi}\left(\frac{1}{4}-\frac{1}{3}\right)=\frac{\mu_0I}{2\pi}\cdot\frac{3-4}{12} \]
\[ B_{net}=-\frac{\mu_0I}{24\pi} \]
Option (B) is twice this value, which results when the factor 2 in $2\pi d$ is dropped.

Final Answer:
$\vec B=-\dfrac{\mu_0I}{24\pi}\hat k$, so option (C) is correct. \[ \boxed{\vec B=-\frac{\mu_0I}{24\pi}\hat k} \]