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: Understanding the Concept:
Each long straight wire produces a magnetic field that circles the wire. Its magnitude at distance \(d\) is \(B=\dfrac{\mu_0I}{2\pi d}\). The direction comes from the right-hand rule. The net field is the vector sum of the two fields.

Step 2: Field of the wire along the x-axis:
The point \((3,4)\) m is at a perpendicular distance \(d_1=4\) m from the x-axis. The current flows along \(+x\), and the point lies on the \(+y\) side of it.
By the right-hand rule, the field direction is \(\hat i\times\hat j=+\hat k\) (out of the page).
\[ \vec B_1=+\frac{\mu_0I}{2\pi(4)}\hat k=\frac{\mu_0I}{8\pi}\hat k \]

Step 3: Field of the wire along the y-axis:
The point is at distance \(d_2=3\) m from the y-axis. The current flows along \(+y\), and the point lies on the \(+x\) side of it.
The direction is \(\hat j\times\hat i=-\hat k\) (into the page).
\[ \vec B_2=-\frac{\mu_0I}{2\pi(3)}\hat k=-\frac{\mu_0I}{6\pi}\hat k \]

Step 4: Add the two fields:
\[ \vec B=\frac{\mu_0I}{2\pi}\left(\frac{1}{4}-\frac{1}{3}\right)\hat k=\frac{\mu_0I}{2\pi}\left(-\frac{1}{12}\right)\hat k=-\frac{\mu_0I}{24\pi}\hat k \]

Step 5: Why the other options are wrong:
Option (A), \(\mu_0I/6\pi\), is only the magnitude of the field of the y-axis wire, with the wrong sign. Option (B), \(-\mu_0I/12\pi\), is twice the correct value. It results if the factor 2 of \(2\pi d\) is dropped. Option (D) has a positive sign, but the wire along the y-axis is nearer and its field wins, so the net field is into the page.

Final Answer:
The net field at the point is \(-\dfrac{\mu_0 I}{24\pi}\hat k\), which is option (C). \[ \boxed{\vec B=-\frac{\mu_0I}{24\pi}\hat k} \]