Unit 03: Magnetic Effects of Current • Chapter 04[cite: 12, 14]
Force on Current Conductor & Parallel Wires
Macroscopic Lorentz integration, mutual wire forces, parallel attraction vs antiparallel repulsion, and the SI Ampere standard.
Electrostatics mein "Like charges repel, Unlike charges attract" universal rule hai. Lekin Magnetism mein ek hair-raising twist aata hai: "Like currents ATTRACT, and Opposite currents REPEL"! Jab do parallel wires mein current same direction mein behta hai, toh dono ke beech magnetic field opposite direction mein hone ke karan cancel hokar kamzor ho jata hai, aur bahar ka high magnetic pressure unhe ek dusre ki taraf push kar deta hai (Attraction). Agar currents opposite hon, toh beech mein magnetic field add hokar super-strong barrier banata hai jo dono wires ko door dhakelta hai (Repulsion)!
1. Force on a Current Conductor
Macroscopic summation of Lorentz forces on all conduction electrons:
Where $\theta$ is the angle between length vector $\vec{L}$ (along current) and $\vec{B}$. Maximum at $\theta = 90^\circ$; zero when parallel ($\theta = 0^\circ$).
2. Force between Two Parallel Wires
Mutual force per unit length between infinite parallel conductors separated by distance $d$:
Obeys Newton's 3rd Law: $\vec{F}_{12} = -\vec{F}_{21}$. SI unit of force per unit length: $\text{N/m}$.
Magnetic Force on a Current-Carrying Conductor ($\vec{F} = I\vec{L}\times\vec{B}$)[cite: 14]
1. Consider a straight segment of conductor of length $L$ and cross-sectional area $A$ carrying steady current $I$ placed in uniform magnetic field $\vec{B}$[cite: 14].
2. Let $n$ be the free electron density[cite: 13]. Total number of conduction electrons in the conductor is $N = n (A L)$[cite: 13].
3. Each electron moving with drift velocity $\vec{v}_d$ experiences magnetic Lorentz force[cite: 13]:
4. Total macroscopic magnetic force $\vec{F}$ is the vector sum over all $N$ free electrons:
5. Since current $I = n e A v_d$ and the direction of current is taken along length vector $\vec{L}$ (opposite to electron drift $\vec{v}_d$)[cite: 13]:
Force Between Two Parallel Current-Carrying Conductors & The Definition of Ampere[cite: 14]
Fig 4.4A: Two parallel conductors carrying currents in the same direction exert mutual attractive forces[cite: 14].
1. Consider two long, straight parallel conductors $1$ and $2$ separated by normal distance $d$ carrying currents $I_1$ and $I_2$ in the same direction[cite: 14].
2. Magnetic field produced by wire $1$ at every point along wire $2$ (directed normally inward $\otimes$ by right-hand rule)[cite: 13]:
3. Wire $2$ carries current $I_2$ in this perpendicular magnetic field $B_1$. By Fleming's Left-Hand Rule, force on length $L$ of wire $2$ acts horizontally towards wire $1$[cite: 14]:
4. Similarly, wire $2$ produces magnetic field $B_2 = \frac{\mu_0 I_2}{2\pi d}$ (outward $\odot$) at wire $1$, exerting equal opposite force $F_{12}$ directed towards wire $2$[cite: 14]:
Official Definition of 1 Ampere (Based on Force between Conductors)[cite: 14]
Set $I_1 = I_2 = 1\text{ A}$, $d = 1\text{ m}$, and $L = 1\text{ m}$ in vacuum:
"One ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed 1 metre apart in vacuum, would produce between these conductors a force equal to $2 \times 10^{-7}\text{ Newton per metre}$ of length."[cite: 14]
Trap 1: Electrostatic vs Magnetic Interaction Duality
• Parallel Beams of Protons in Vacuum: Repel each other! Electrostatic Coulomb repulsion ($F_e = \frac{kq_1q_2}{r^2}$) is much stronger than magnetic attraction ($F_m = \frac{v^2}{c^2} F_e$).
• Parallel Current-Carrying Copper Wires: ATTRACT each other[cite: 14]! Wires are overall electrically neutral (positive lattice ions cancel electron charge $\implies F_e = 0$), so pure magnetic attraction dominates[cite: 14].
Trap 2: The "Unequal Currents" Force Trap
*Question:* If Wire 1 carries $100\text{ A}$ and Wire 2 carries $1\text{ A}$, which wire experiences the greater force?
*Trap Answer:* Wire 2 feels 100 times more force.
*Correct Answer:* BOTH EXPERIENCE EXACTLY THE SAME FORCE!
$F_{12} = F_{21} = \frac{\mu_0 I_1 I_2 L}{2\pi d}$[cite: 14]. Newton's Third Law is strictly obeyed[cite: 14].
Trap 3: Arbitrary Shaped Wire in Uniform Field
For any curved wire of arbitrary shape carrying current $I$ in a uniform magnetic field $\vec{B}$: $$\vec{F} = \int I (d\vec{l} \times \vec{B}) = I \left( \oint d\vec{l} \right) \times \vec{B} = I (\vec{L}' \times \vec{B})$$ Where $\vec{L}'$ is the direct vector displacement from the start point to the end point! For any closed planar loop of any shape in a uniform field, $\oint d\vec{l} = 0 \implies \vec{F}_{\text{net}} = 0$.
Trap 4: Direction of Force (Fleming's Left-Hand Rule)
Stretch Forefinger (Field $\vec{B}$), Middle finger (Current $I$), and Thumb: Thumb gives Force $\vec{F}$.
Remember: Current in Wire 2 is in the field of Wire 1 ($B_1$), NOT in its own field $B_2$! A wire cannot exert a net magnetic force on itself.
Problem: A horizontal wire $A$ carries a current of $50\text{ A}$ and is rigidly fixed. Another wire $B$ is placed vertically below and parallel to wire $A$ at a distance of $2.0\text{ mm}$. If the mass per unit length of wire $B$ is $2.5 \times 10^{-3}\text{ kg/m}$, find the magnitude and direction of the current required in wire $B$ such that it remains magnetically suspended in air. (Take $g = 9.8\text{ m/s}^2$).
Solution:
1. Wire $B$ experiences gravitational downward force per unit length: $(F_g / L) = \lambda g = (2.5 \times 10^{-3}) \times 9.8 = 2.45 \times 10^{-2}\text{ N/m}$.
2. For wire $B$ to remain suspended, the magnetic force per unit length must be directed vertically upward (Attractive force towards wire $A$).
3. Attractive force requires currents in both wires to flow in the SAME DIRECTION[cite: 14].
4. Equating upward magnetic force to downward gravity[cite: 14]:
5. Simplify and solve for $I_B$:
Verdict: $I_B = 4.9\text{ A}$ flowing parallel (in the same direction) as in wire $A$.
Problem: A rectangular loop of dimensions $a \times b$ carrying current $I_2$ is placed parallel to a long straight wire carrying current $I_1$ in the same plane, with the nearer long side of length $a$ at distance $r$ from the straight wire. Calculate the net magnetic force on the rectangular loop.
Solution:
1. Let the two sides parallel to the wire be at distances $r$ (nearer side) and $(r + b)$ (farther side), both of length $a$.
2. The two perpendicular sides of length $b$ experience equal and opposite magnetic forces by symmetry ($\vec{F}_{\text{top}} + \vec{F}_{\text{bottom}} = 0$), canceling out completely.
3. Force on nearer side (current parallel to $I_1$ $\implies$ Attractive towards wire)[cite: 14]:
4. Force on farther side (current antiparallel to $I_1$ $\implies$ Repulsive away from wire)[cite: 14]:
5. Net force on the loop ($F_1 > F_2$):
Conductor Force
$\vec{F} = I(\vec{L}\times\vec{B})$[cite: 14]
Parallel Wires Force
$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}$[cite: 14]
Current Direction Rule
Same $\to$ Attract; Opp $\to$ Repel[cite: 14]
Definition of Ampere
$2 \times 10^{-7}\text{ N/m}$ at $1\text{ m}$[cite: 14]