Question

The figure given below shows three straight long parallel conductors kept in x-y plane, carrying currents 2I, I and 3I respectively as shown in figure. Find the magnitude and direction of net magnetic field at a point on conductor 1.

Hint:

Always assign vector unit vectors: $+\hat{i}$ along wire 1, $-\hat{j}$ downwards along y-axis, and $\hat{k}$ perpendicular to x-y plane. Applying $\vec{B} = \frac{\mu_0 I}{2\pi r} (\hat{d}_I \times \hat{r}_p)$ gives unambiguous signs.

Answer 1

Concept:
• Magnetic field at distance $r$ from an infinitely long straight wire carrying current $I$ is $B = \frac{\mu_0 I}{2\pi r}$.

• Direction of magnetic field is determined by Right Hand Thumb Rule.

• Net magnetic field is the vector sum of individual magnetic fields produced by surrounding conductors.

Step 1: Determine magnetic field due to Conductor 2
Conductor 2 carries current $I$ in the $+x$ direction at distance $d$ below conductor 1.
By Right Hand Thumb Rule, magnetic field $\vec{B}_2$ at conductor 1 points out of the page ($+z$ direction, along $+\hat{k}$):
\[ \vec{B}_2 = \frac{\mu_0 I}{2\pi d} \hat{k} \]

Step 2: Determine magnetic field due to Conductor 3
Conductor 3 carries current $3I$ in the $-x$ direction at distance $2d$ below conductor 1.
By Right Hand Thumb Rule, magnetic field $\vec{B}_3$ at conductor 1 points into the page ($-z$ direction, along $-\hat{k}$):
\[ \vec{B}_3 = \frac{\mu_0 (3I)}{2\pi (2d)} (-\hat{k}) = -\frac{3 \mu_0 I}{4\pi d} \hat{k} \]

Step 3: Calculate net magnetic field at conductor 1
\[ \vec{B}_{net} = \vec{B}_2 + \vec{B}_3 \]
\[ \vec{B}_{net} = \frac{\mu_0 I}{2\pi d} \hat{k} - \frac{3 \mu_0 I}{4\pi d} \hat{k} \]
Take common denominator $4\pi d$:
\[ \vec{B}_{net} = \left( \frac{2 \mu_0 I - 3 \mu_0 I}{4\pi d} \right) \hat{k} = -\frac{\mu_0 I}{4\pi d} \hat{k} \]

Step 4: Conclusion
Magnitude of net magnetic field at conductor 1 is $B_{net} = \frac{\mu_0 I}{4\pi d}$, directed into the plane of the paper (along the negative z-axis, $-\hat{k}$).