The figure given below shows three straight long parallel conductors 1, 2 and 3 kept in x-y plane, carrying currents 2I, I and 3I respectively as shown in figure. Find the magnitude and direction of :
(i) net magnetic field at a point on conductor 1 and
(ii) net magnetic force acting on unit length of conductor 1, due to conductors 2 and 3.
Show Hint
For parallel wires, remember the simple rule: Like currents attract, unlike currents repel. Conductor 1 and 2 carry currents in the same direction, so 2 attracts 1 upwards ($+\hat{j}$). Conductor 1 and 3 carry opposite currents, so 3 repels 1 upwards ($+\hat{j}$). Simply add their scalar forces together!
Concept: • Magnetic Field due to a Long Straight Wire: The magnitude of the magnetic field $B$ at a perpendicular distance $d$ from an infinitely long straight conductor carrying current $I$ is given by Ampere's Law:
\[
B = \frac{\mu_0 I}{2\pi d}
\]
The direction of this magnetic field is determined by the Right-Hand Thumb Rule.
• Magnetic Force between Parallel Wires: The magnetic force per unit length ($F/L$) experienced by a conductor carrying current $I_1$ due to another parallel conductor carrying current $I_2$ separated by a distance $d$ is:
\[
\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}
\]
Parallel currents flowing in the same direction attract each other, whereas currents flowing in opposite directions repel each other.
Part (i): Net magnetic field at a point on conductor 1.
Conductor 1 is located at the top. The magnetic field at any point on conductor 1 is the vector sum of the fields produced by conductor 2 and conductor 3.
• Field due to Conductor 2 ($B_2$): Conductor 2 is at a distance $d$ below conductor 1 and carries a current $I$ in the $+x$ direction. Applying the Right-Hand Thumb Rule, pointing the thumb along $+x$ causes the fingers to curl out of the page ($+z$ direction) at the position of conductor 1.
\[
\vec{B}_2 = \frac{\mu_0 I}{2\pi d} \hat{k}
\]
• Field due to Conductor 3 ($B_3$): Conductor 3 is at a total distance of $2d$ below conductor 1 and carries a current $3I$ in the $-x$ direction. Pointing the right thumb along $-x$ causes the fingers to curl into the page ($-z$ direction) at the position of conductor 1.
\[
\vec{B}_3 = -\frac{\mu_0 (3I)}{2\pi (2d)} \hat{k} = -\frac{3\mu_0 I}{4\pi d} \hat{k}
\]
The net magnetic field $\vec{B}_{\text{net}}$ is the vector sum of these two components:
\[
\vec{B}_{\text{net}} = \vec{B}_2 + \vec{B}_3 = \frac{\mu_0 I}{2\pi d} \hat{k} - \frac{3\mu_0 I}{4\pi d} \hat{k}
\]
Taking a common denominator of $4\pi d$:
\[
\vec{B}_{\text{net}} = \left( \frac{2\mu_0 I - 3\mu_0 I}{4\pi d} \right) \hat{k} = -\frac{\mu_0 I}{4\pi d} \hat{k}
\]
Thus, the magnitude of the net magnetic field is $\frac{\mu_0 I}{4\pi d}$ and its direction is pointing into the plane of the page ($-\hat{k}$).
Part (ii): Net magnetic force acting on unit length of conductor 1.
We can calculate the force per unit length using the magnetic force equation $\vec{F}/L = \vec{I} \times \vec{B}_{\text{net}}$:
The current vector for unit length of conductor 1 is:
\[
\vec{I}_1 = 2I \hat{i}
\]
Using the net magnetic field calculated in part (i):
\[
\frac{\vec{F}}{L} = \vec{I}_1 \times \vec{B}_{\text{net}} = (2I \hat{i}) \times \left( -\frac{\mu_0 I}{4\pi d} \hat{k} \right)
\]
Factoring out the scalar components:
\[
\frac{\vec{F}}{L} = - \frac{2\mu_0 I^2}{4\pi d} (\hat{i} \times \hat{k})
\]
Since $\hat{i} \times \hat{k} = -\hat{j}$:
\[
\frac{\vec{F}}{L} = - \frac{\mu_0 I^2}{2\pi d} (-\hat{j}) = \frac{\mu_0 I^2}{2\pi d} \hat{j}
\]
Thus, the net force per unit length has a magnitude of $\frac{\mu_0 I^2}{2\pi d}$ and is directed vertically upwards ($+\hat{j}$ direction).
