Magnetism - Magnetic Force on Currents Solved Examples (Class 12 Physics)
Calculate magnetic forces on current-carrying conductors using Lorentz force law. These problems cover force on straight wires, loops, and moving charges i
TL;DR: Calculate magnetic forces on current-carrying conductors using Lorentz force law. These problems cover force on straight wires, loops, and moving char…
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
Calculate magnetic forces on current-carrying conductors using Lorentz force law. These problems cover force on straight wires, loops, and moving charges i
Magnetism - Magnetic Force on Currents — Solved Numerical Examples (Step by Step)
Example 1: A straight conductor of length 2 m carries a current of 5 A perpendicular to a magnetic field of 0.4 T. Find the force on the conductor.
Solution: Using F = BIL (when angle = 90°)
F = 0.4 × 5 × 2 = 4 N
Example 2: A current-carrying conductor in a magnetic field experiences a force. If the conductor is at 30° to the field, with B = 2 T, I = 3 A, and L = 1 m, find the force.
Solution: Using F = BIL sin(θ)
F = 2 × 3 × 1 × sin(30°) = 2 × 3 × 1 × 0.5 = 3 N
Example 3: An electron moves with velocity 2 × 10^6 m/s perpendicular to a magnetic field of 0.5 T. Find the magnetic force on the electron. (Charge of electron = 1.6 × 10^-19 C)
Solution: Using F = qvB (when velocity is perpendicular to field)
F = 1.6 × 10^-19 × 2 × 10^6 × 0.5
F = 1.6 × 10^-19 × 10^6 = 1.6 × 10^-13 N
Example 4: A rectangular loop with dimensions 0.3 m × 0.2 m carries a current of 4 A. It is placed in a uniform magnetic field of 0.5 T perpendicular to the plane. Find the torque on the loop.
Solution: Magnetic moment M = IA = 4 × (0.3 × 0.2) = 4 × 0.06 = 0.24 A·m²
Torque τ = MB sin(θ) = 0.24 × 0.5 × sin(90°) = 0.24 × 0.5 × 1 = 0.12 N·m
Example 5: Two parallel wires separated by 10 cm carry currents of 5 A and 3 A in the same direction. Calculate the force per unit length between them. (μ₀ = 4π × 10^-7 T·m/A)
Solution: Force per unit length between parallel wires: F/L = (μ₀ I₁ I₂) / (2π d)
F/L = (4π × 10^-7 × 5 × 3) / (2π × 0.1)
F/L = (4 × 10^-7 × 15) / (0.2) = (60 × 10^-7) / 0.2 = 3 × 10^-5 N/m
Example 6: A proton moves with velocity 3 × 10^7 m/s at 60° to a magnetic field of 1.2 T. Find the magnetic force. (Charge of proton = 1.6 × 10^-19 C)
Solution: Using F = qvB sin(θ)
F = 1.6 × 10^-19 × 3 × 10^7 × 1.2 × sin(60°)
F = 1.6 × 10^-19 × 3 × 10^7 × 1.2 × (√3/2)
F = 1.6 × 10^-19 × 3 × 10^7 × 1.2 × 0.866
F = 1.6 × 3 × 1.2 × 0.866 × 10^-12 = 4.98 × 10^-12 N
Tips
- Magnetic force on a charge: F = qvB sin(θ); it is perpendicular to both velocity and field (use right-hand rule).
- Force on a current-carrying conductor: F = BIL sin(θ).
- Torque on a loop: τ = NIAB sin(θ), where N is number of turns, M = NIA is magnetic moment.
Frequently Asked Questions
Why does a magnetic field not do work on a moving charge?
Magnetic force is always perpendicular to velocity, so F·v = 0. Since work W = F·d and d is parallel to v, the magnetic force does no work.
What determines the direction of force on a current-carrying conductor in a magnetic field?
Use the right-hand rule (or left-hand rule): point fingers in current direction, curl them toward the field direction; thumb points in force direction.
More Physics Solved Examples
🤖 Stuck on any of these? Ask Syllab's free AI Tutor to explain step by step →