Force on Current-Carrying Conductors: Electromagnetism in Action
Movie Hook: Iron Man’s Repulsor Technology
In Iron Man, Tony Stark’s repulsor beams use electromagnetic forces to push and pull objects. While Hollywood takes creative liberty, the real physics is equally impressive! When electric currents flow through conductors in magnetic fields, they experience forces - the same principle behind electric motors, maglev trains, and loudspeakers. Master this, and you’ll understand how we convert electrical energy into mechanical motion!
The Big Picture
What’s the Deal?
- Current-carrying wires in magnetic fields experience forces
- This is the basis for electric motors and generators
- Parallel wires carrying currents attract or repel each other
- Torque on current loops is how galvanometers work
JEE Perspective:
- JEE Main: Force on straight wires, parallel conductors, rectangular loops
- JEE Advanced: Torque calculations, work done, non-uniform fields, complex geometries
Core Concept: Force on Current Element
The Fundamental Law
A current element $I\,d\vec{l}$ in magnetic field $\vec{B}$ experiences force:
$$\boxed{d\vec{F} = I\,d\vec{l} \times \vec{B}}$$Magnitude:
$$\boxed{dF = I\,dl\,B\sin\theta}$$where $\theta$ is the angle between current direction and magnetic field
For straight conductor of length L:
$$\boxed{\vec{F} = I\vec{L} \times \vec{B}}$$Magnitude:
$$\boxed{F = BIL\sin\theta}$$Maximum force: When current ⊥ field (θ = 90°): $F_{\max} = BIL$
Direction: Right-hand rule
- Fingers: Current direction (I)
- Bend toward: Magnetic field (B)
- Thumb: Force direction (F)
Memory Trick: “FBI - Force, B-field, I-current”
- Point fingers along I
- Bend into B
- Thumb shows F
Understanding the Force: Microscopic View
Why does wire experience force?
Current = Flow of charges with drift velocity $v_d$
Number of charge carriers in length L:
$$N = nAL$$where n = number density, A = cross-sectional area
Force on each charge:
$$f = qv_dB$$Total force:
$$F = Nf = (nAL)(qv_dB)$$Since current $I = nAqv_d$:
$$\boxed{F = BIL}$$Key Insight: Macroscopic force on wire = Sum of microscopic forces on all moving charges!
Standard Configurations
1. Straight Wire in Uniform Field
Setup: Wire of length L carrying current I perpendicular to field B
Force:
$$\boxed{F = BIL}$$Direction: Perpendicular to both I and B (right-hand rule)
Applications:
- DC motors (force on armature wires)
- Electromagnetic railguns
- Loudspeaker coils
2. Two Parallel Conductors
Setup: Two long parallel wires separated by distance d, carrying currents I₁ and I₂
Magnetic field from wire 1 at wire 2:
$$B_1 = \frac{\mu_0 I_1}{2\pi d}$$Force on length L of wire 2:
$$\boxed{F = \frac{\mu_0 I_1 I_2 L}{2\pi d}}$$Force per unit length:
$$\boxed{\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}}$$Direction:
- Same direction currents: Attract (force inward)
- Opposite direction currents: Repel (force outward)
Memory Trick: “Same currents are friends - they ATTRACT!”
Interactive Demo: Forces Between Conductors
Watch how parallel conductors attract or repel based on current direction.
Definition of Ampere (SI unit): The ampere is defined such that two parallel conductors 1 meter apart, each carrying 1 ampere, experience a force of exactly $2 \times 10^{-7}$ N per meter.
$$\frac{F}{L} = \frac{4\pi \times 10^{-7} \times 1 \times 1}{2\pi \times 1} = 2 \times 10^{-7} \text{ N/m}$$3. Rectangular Loop in Magnetic Field
Setup: Rectangular loop (sides a × b) in uniform field B, current I
Forces on each side:
- Two sides parallel to B: $F = 0$ (θ = 0°)
- Two sides perpendicular to B: $F = BIa$ (opposite directions)
Net force: Zero (forces cancel)
BUT… there’s a TORQUE!
$$\boxed{\tau = BINA\sin\theta}$$where:
- N = number of turns
- A = area of loop
- θ = angle between normal to loop and B
For single turn: $\tau = BIA\sin\theta$
Maximum torque: When loop plane is parallel to B (θ = 90°)
$$\tau_{\max} = BIA$$Stable equilibrium: When loop plane is perpendicular to B (θ = 0°)
4. Circular Loop in Uniform Field
Same torque equation:
$$\boxed{\tau = BINA\sin\theta}$$where $A = \pi r^2$ for radius r
Special property: Torque is same at all orientations for given θ!
Magnetic Dipole Moment
Definition:
$$\boxed{\vec{M} = NI\vec{A}}$$where $\vec{A}$ is area vector (perpendicular to plane, right-hand rule for current)
Magnitude: $M = NIA$
Torque in terms of dipole moment:
$$\boxed{\vec{\tau} = \vec{M} \times \vec{B}}$$Magnitude: $\tau = MB\sin\theta$
Potential energy:
$$\boxed{U = -\vec{M} \cdot \vec{B} = -MB\cos\theta}$$Energy configurations:
- Minimum U: θ = 0° (stable, M ∥ B)
- Maximum U: θ = 180° (unstable, M anti-parallel B)
Work done to rotate from θ₁ to θ₂:
$$\boxed{W = MB(\cos\theta_1 - \cos\theta_2)}$$Memory Trick: “Magnetic dipole acts like electric dipole in E-field!”
- Electric: $\tau = pE\sin\theta$, $U = -pE\cos\theta$
- Magnetic: $\tau = MB\sin\theta$, $U = -MB\cos\theta$
Interactive Demo: Forces Between Wires
import numpy as np
import matplotlib.pyplot as plt
def force_per_length(I1, I2, d):
"""
Force per unit length between parallel wires
I1, I2: currents (A)
d: separation (m)
Returns: Force per meter (N/m)
"""
mu_0 = 4 * np.pi * 1e-7
return (mu_0 * I1 * I2) / (2 * np.pi * d)
# Explore how force varies with distance
distances = np.linspace(0.01, 0.5, 100) # 1 cm to 50 cm
I1 = 10 # Amperes
I2 = 10 # Amperes
forces = force_per_length(I1, I2, distances)
plt.figure(figsize=(12, 5))
# Plot 1: Force vs Distance
plt.subplot(1, 2, 1)
plt.plot(distances * 100, forces * 1e3, linewidth=2.5, color='darkred')
plt.xlabel('Distance (cm)', fontsize=12)
plt.ylabel('Force per meter (mN/m)', fontsize=12)
plt.title(f'Force Between Parallel Wires (I₁=I₂={I1}A)', fontsize=13, fontweight='bold')
plt.grid(True, alpha=0.3)
plt.axhline(y=0, color='k', linewidth=0.5)
# Annotations
plt.annotate('Force ∝ 1/d', xy=(25, forces[50]*1e3), fontsize=11,
bbox=dict(boxstyle='round', facecolor='yellow', alpha=0.7))
# Plot 2: Force vs Current (at fixed distance)
plt.subplot(1, 2, 2)
d_fixed = 0.1 # 10 cm
currents = np.linspace(1, 20, 100)
forces_I = force_per_length(currents, currents, d_fixed)
plt.plot(currents, forces_I * 1e3, linewidth=2.5, color='darkblue')
plt.xlabel('Current in each wire (A)', fontsize=12)
plt.ylabel('Force per meter (mN/m)', fontsize=12)
plt.title(f'Force vs Current (d={d_fixed*100}cm)', fontsize=13, fontweight='bold')
plt.grid(True, alpha=0.3)
plt.annotate('Force ∝ I²', xy=(10, forces_I[50]*1e3), fontsize=11,
bbox=dict(boxstyle='round', facecolor='lightblue', alpha=0.7))
plt.tight_layout()
plt.show()
# Calculate some specific values
print("Example Calculations:")
print(f"At d=5cm, I=10A each: F/L = {force_per_length(10, 10, 0.05)*1e3:.3f} mN/m")
print(f"At d=10cm, I=10A each: F/L = {force_per_length(10, 10, 0.10)*1e3:.3f} mN/m")
print(f"At d=1m, I=1A each: F/L = {force_per_length(1, 1, 1)*1e7:.3f} × 10⁻⁷ N/m")
print(f"(Should be 2×10⁻⁷ N/m for SI definition): {2e-7} N/m ✓")
Observe: Force decreases with distance (∝ 1/d) and increases with current product (∝ I₁I₂)
Applications: Real-World Devices
1. Moving Coil Galvanometer
Principle: Torque on current-carrying coil in magnetic field
Components:
- Coil of N turns, area A
- Radial magnetic field B (uniform)
- Restoring spring with constant k
Torque due to magnetic field:
$$\tau_B = NIBA$$Restoring torque from spring:
$$\tau_s = k\theta$$At equilibrium:
$$NIBA = k\theta$$ $$\boxed{\theta = \frac{NIBA}{k} = \frac{NAB}{k} \cdot I}$$Current sensitivity:
$$\boxed{S_I = \frac{\theta}{I} = \frac{NAB}{k}}$$To increase sensitivity:
- Increase N (more turns)
- Increase A (larger coil area)
- Increase B (stronger magnet)
- Decrease k (weaker spring)
2. Conversion to Ammeter
Problem: Galvanometer has small current capacity (typically μA to mA)
Solution: Connect low-resistance shunt in parallel
Shunt resistance:
$$\boxed{S = \frac{I_g G}{I - I_g}}$$where:
- I_g = full-scale deflection current of galvanometer
- G = galvanometer resistance
- I = desired full-scale current
Key: Shunt carries most of the current, galvanometer carries small fraction
3. Conversion to Voltmeter
Solution: Connect high resistance in series
Series resistance:
$$\boxed{R = \frac{V}{I_g} - G}$$where V is desired full-scale voltage
Key: High resistance limits current through galvanometer
4. DC Motor
Principle: Torque on coil rotates armature
Components:
- Armature coil in magnetic field
- Commutator (reverses current every half turn)
- Brushes (maintain electrical contact)
Torque: $\tau = NIBA\sin\theta$
Commutator ensures: Torque always in same direction (continuous rotation)
Power: $P = \tau \omega = (BIA)I\omega$
Common JEE Mistakes (Avoid These!)
Mistake 1: Wrong Direction (Right-Hand Rule)
Wrong: Force is along current direction Right: Force is perpendicular to BOTH current and field (use FBI rule!)
Mistake 2: Confusing Attraction/Repulsion
Wrong: Opposite currents attract Right: SAME direction currents attract, OPPOSITE currents repel!
Mistake 3: Force on Bent Wire
Wrong: Calculating force for each segment separately Right: Effective length is straight-line distance between ends!
Example: Semicircular wire in uniform field
- Effective length = diameter = 2R
- F = BI(2R), not the arc length!
Mistake 4: Torque at Equilibrium
Wrong: Torque is maximum at equilibrium (θ = 0°) Right: Torque is ZERO at equilibrium! Maximum at θ = 90°
Mistake 5: Magnetic Dipole Energy
Wrong: U = MB cos θ (positive) Right: U = -MB cos θ (negative sign is crucial!)
Problem-Solving Strategy: The LIFT Method
Length and current identification Identify field direction and magnitude Force direction (right-hand rule) Torque if there’s a moment arm
Practice Problems
Level 1: JEE Main Foundations
Problem 1.1: A wire of length 50 cm carrying a current of 2 A is placed perpendicular to a magnetic field of 0.5 T. Find the force on the wire.
Solution
Given:
- L = 0.5 m
- I = 2 A
- B = 0.5 T
- θ = 90° (perpendicular)
Formula: $F = BIL\sin\theta$
$$F = 0.5 \times 2 \times 0.5 \times \sin 90°$$ $$F = 0.5 \times 2 \times 0.5 \times 1 = 0.5 \text{ N}$$Answer: 0.5 N (perpendicular to both wire and field)
Problem 1.2: Two parallel wires 10 cm apart carry currents of 5 A each in the same direction. Find the force per unit length between them.
Solution
Given:
- d = 0.1 m
- I₁ = I₂ = 5 A
- Same direction (attractive)
Formula: $\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}$
$$\frac{F}{L} = \frac{4\pi \times 10^{-7} \times 5 \times 5}{2\pi \times 0.1}$$ $$\frac{F}{L} = \frac{100 \times 10^{-7}}{0.2} = 5 \times 10^{-5} \text{ N/m}$$ $$\frac{F}{L} = 50 \text{ μN/m}$$Answer: 50 μN/m (attractive force)
Problem 1.3: A rectangular coil of 100 turns, area 0.02 m², carries a current of 0.5 A in a magnetic field of 0.8 T. Find the maximum torque.
Solution
Given:
- N = 100 turns
- A = 0.02 m²
- I = 0.5 A
- B = 0.8 T
Maximum torque (when θ = 90°):
$$\tau_{\max} = NIBA$$ $$\tau = 100 \times 0.5 \times 0.8 \times 0.02$$ $$\tau = 0.8 \text{ N·m}$$Answer: 0.8 N·m
Level 2: JEE Advanced Application
Problem 2.1: A semicircular wire of radius 10 cm carrying 5 A is placed in a uniform field of 0.5 T perpendicular to the plane of the wire. Find the force on the wire.
Solution
Key Insight: For any curved wire in uniform field, effective length is the straight-line distance between end points!
Given:
- Radius r = 0.1 m
- I = 5 A
- B = 0.5 T
- Perpendicular to plane
Effective length: Diameter = 2r = 0.2 m
Force:
$$F = BIL_{\text{eff}} = 0.5 \times 5 \times 0.2$$ $$F = 0.5 \text{ N}$$Direction: From one end to the other, along the diameter
Answer: 0.5 N (along the diameter)
JEE Trick: Arc length πr doesn’t matter - only straight-line distance!
Problem 2.2: A galvanometer has 50 turns, area 10 cm², and experiences a magnetic field of 0.2 T. The spring constant is 10⁻⁴ N·m/rad. Find the current for 30° deflection.
Solution
Given:
- N = 50
- A = 10 cm² = 10 × 10⁻⁴ m² = 10⁻³ m²
- B = 0.2 T
- k = 10⁻⁴ N·m/rad
- θ = 30° = π/6 rad
At equilibrium:
$$NIBA = k\theta$$ $$I = \frac{k\theta}{NBA}$$ $$I = \frac{10^{-4} \times (\pi/6)}{50 \times 0.2 \times 10^{-3}}$$ $$I = \frac{10^{-4} \times 0.524}{10^{-2}} = \frac{5.24 \times 10^{-5}}{10^{-2}}$$ $$I = 5.24 \times 10^{-3} \text{ A} = 5.24 \text{ mA}$$Answer: 5.24 mA
Problem 2.3: A galvanometer of resistance 100 Ω gives full-scale deflection for 1 mA. How would you convert it to: (a) an ammeter of range 0-1 A, (b) a voltmeter of range 0-10 V?
Solution
Given:
- G = 100 Ω
- I_g = 1 mA = 10⁻³ A
(a) Ammeter (0-1 A):
Need shunt resistance S in parallel:
$$S = \frac{I_g G}{I - I_g} = \frac{10^{-3} \times 100}{1 - 10^{-3}}$$ $$S = \frac{0.1}{0.999} = 0.1001 \text{ Ω} \approx 0.1 \text{ Ω}$$Configuration: 0.1 Ω shunt in parallel with galvanometer
(b) Voltmeter (0-10 V):
Need series resistance R:
$$R = \frac{V}{I_g} - G = \frac{10}{10^{-3}} - 100$$ $$R = 10,000 - 100 = 9,900 \text{ Ω} = 9.9 \text{ kΩ}$$Configuration: 9.9 kΩ resistor in series with galvanometer
Answers:
- (a) Shunt of 0.1 Ω in parallel
- (b) Resistance of 9.9 kΩ in series
Level 3: JEE Advanced Mastery
Problem 3.1: Three parallel wires are located at the corners of an equilateral triangle of side 10 cm. Each carries a current of 5 A in the same direction (perpendicular to plane, coming out). Find the force on each wire per unit length.
Solution
Setup: Equilateral triangle, side a = 0.1 m, all currents equal I = 5 A
Force on wire 1 due to wire 2:
$$F_{12} = \frac{\mu_0 I^2}{2\pi a}$$(attractive, toward wire 2)
Force on wire 1 due to wire 3:
$$F_{13} = \frac{\mu_0 I^2}{2\pi a}$$(attractive, toward wire 3)
Angle between F₁₂ and F₁₃: 60° (equilateral triangle)
Magnitude of each force:
$$F_{12} = F_{13} = \frac{4\pi \times 10^{-7} \times 25}{2\pi \times 0.1} = \frac{10^{-5}}{0.1} = 10^{-4} \text{ N/m}$$Resultant force:
$$F_{\text{net}} = \sqrt{F_{12}^2 + F_{13}^2 + 2F_{12}F_{13}\cos 60°}$$ $$F_{\text{net}} = F_{12}\sqrt{1 + 1 + 2 \times 0.5} = F_{12}\sqrt{3}$$ $$F_{\text{net}} = 10^{-4} \times \sqrt{3} = 1.73 \times 10^{-4} \text{ N/m}$$Direction: Toward the centroid of the triangle (bisecting the 60° angle)
Answer: $1.73 \times 10^{-4}$ N/m = 173 μN/m toward the center
By symmetry: All three wires experience same magnitude force toward center
Problem 3.2: A circular coil of radius 5 cm with 20 turns carries a current of 1 A. It is placed in a magnetic field of 0.5 T such that the plane of the coil makes 60° with the field. Find: (a) the torque, (b) work done to rotate it to perpendicular position.
Solution
Given:
- r = 0.05 m, so A = πr² = π(0.05)² = 7.85 × 10⁻³ m²
- N = 20
- I = 1 A
- B = 0.5 T
Initial configuration: Plane makes 60° with B
- This means normal to plane makes 30° with B (complementary angle)
- So θ₁ = 30°
Final configuration: Plane perpendicular to B
- Normal parallel to B
- θ₂ = 0°
(a) Initial torque:
$$\tau = NIBA\sin\theta_1$$ $$\tau = 20 \times 1 \times 0.5 \times 7.85 \times 10^{-3} \times \sin 30°$$ $$\tau = 10 \times 7.85 \times 10^{-3} \times 0.5$$ $$\tau = 0.0393 \text{ N·m} = 39.3 \text{ mN·m}$$(b) Work done:
$$W = MB(\cos\theta_1 - \cos\theta_2)$$where M = NIA = 20 × 1 × 7.85 × 10⁻³ = 0.157 A·m²
$$W = 0.157 \times 0.5 \times (\cos 30° - \cos 0°)$$ $$W = 0.0785 \times (0.866 - 1)$$ $$W = 0.0785 \times (-0.134) = -0.0105 \text{ J}$$Negative work means field does the work (rotation is in favored direction)
If we do the work against the field:
$$W_{\text{external}} = +0.0105 \text{ J} = 10.5 \text{ mJ}$$Answers:
- (a) Torque = 39.3 mN·m
- (b) Work by field = -10.5 mJ (or external work = +10.5 mJ)
Problem 3.3: A conducting rod of length L and mass m is suspended by two flexible wires in a magnetic field B perpendicular to the rod. When current I flows through the rod, it makes angle θ with vertical at equilibrium. Find the current I.
Solution
Setup:
- Rod suspended by wires, can swing like a pendulum
- Magnetic force acts horizontally (perpendicular to both I and B)
- At equilibrium: balances component of gravity
Forces:
- Weight: mg (downward)
- Tension: T (along wires)
- Magnetic force: F = BIL (horizontal)
At equilibrium angle θ:
Horizontal component: $T\sin\theta = BIL$
Vertical component: $T\cos\theta = mg$
Dividing:
$$\tan\theta = \frac{BIL}{mg}$$ $$\boxed{I = \frac{mg\tan\theta}{BL}}$$Answer: $I = \frac{mg\tan\theta}{BL}$
Physical interpretation:
- Larger current → larger angle (stronger magnetic force)
- Heavier rod → larger current needed for same angle
- Stronger field → smaller current needed
Special Case: Effective Length
Important Theorem: For any shape of wire in a uniform magnetic field, the force equals the force on a straight wire connecting the end points.
Mathematical form:
$$\vec{F} = I\vec{L}_{\text{eff}} \times \vec{B}$$where $\vec{L}_{\text{eff}}$ is the vector from start to end point.
Applications:
- Semicircle: $L_{\text{eff}} = 2r$ (diameter)
- Quarter circle: $L_{\text{eff}} = r\sqrt{2}$ (diagonal)
- Closed loop: $L_{\text{eff}} = 0$, so F = 0!
JEE Gold: Closed loop in uniform field has ZERO net force (but can have torque!)
Connections: The Web of Knowledge
Link to Lorentz Force
- Force on current = Sum of forces on moving charges
- F = BIL derived from F = qvB
- See: Force on Moving Charge
Link to Electromagnetic Induction
- Moving conductor in B generates EMF (motional EMF)
- Force and EMF are related through power
- See: Motional EMF
Link to Galvanometer
- Torque on coil is the basis for measurement
- Moving coil galvanometer uses this principle
- See: Moving Coil Galvanometer
Link to Electric Motors
- Continuous torque from commutator reversal
- Conversion of electrical to mechanical energy
- See: DC Motors
Quick Revision: Formula Sheet
| Quantity | Formula | Note |
|---|---|---|
| Force on wire | $F = BIL\sin\theta$ | Max when I ⊥ B |
| Force between wires | $\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}$ | Attractive if same direction |
| Torque on coil | $\tau = NIBA\sin\theta$ | Max when plane ∥ B |
| Magnetic moment | $M = NIA$ | Current × Area |
| Potential energy | $U = -MB\cos\theta$ | Min when M ∥ B |
| Work done | $W = MB(\cos\theta_1 - \cos\theta_2)$ | Rotation |
| Ammeter shunt | $S = \frac{I_g G}{I - I_g}$ | Parallel, low R |
| Voltmeter series R | $R = \frac{V}{I_g} - G$ | Series, high R |
Remember:
- Same currents attract, opposite currents repel
- Effective length for curved wire = straight-line distance
- Closed loop: F = 0, but τ may not be zero!
Exam Strategy
Pattern Recognition:
- “Two parallel wires” → Force per unit length formula
- “Coil in magnetic field” → Torque or energy
- “Galvanometer converted to” → Shunt or series resistance
Time Savers:
- Right-hand rule: Practice until automatic!
- For bent wires: Use effective length theorem
- Check units: Force in N, torque in N·m
Common Traps:
- Direction of current (arrow carefully!)
- Angle confusion: plane vs normal
- Sign of work: who does the work?
Summary: Key Takeaways
Force on Current: $\vec{F} = I\vec{L} \times \vec{B}$ - perpendicular to both I and B
Parallel Conductors: Same currents attract ($\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}$), opposite repel
Torque on Loop: $\tau = NIBA\sin\theta$ - maximum when plane parallel to B
Magnetic Dipole: Moment M = NIA, behaves like electric dipole (U = -MB cos θ)
Galvanometer: Converts to ammeter (low R shunt) or voltmeter (high R series)
Effective Length: Force on curved wire = force on straight connector
JEE Pattern: 50% torque/energy, 30% parallel wires, 20% galvanometer conversion
Last Updated: March 25, 2025 Next Topic: Moving Coil Galvanometer Previous Topic: Force on Moving Charge