krit.club logo

Moving Charges and Magnetism - Force between Two Parallel Currents

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Magnetic field produced by a long straight wire carrying current II at a distance dd is given by B=μ0I2πdB = \frac{\mu_0 I}{2\pi d} based on Ampere's Circuital Law.

•

A current-carrying conductor of length LL placed in a magnetic field BB experiences a Lorentz force: F=ILBsin⁡θF = I L B \sin \theta.

•

For two parallel wires, the magnetic field of the first wire (B1B_1) exerts a force on the second wire (I2I_2). The force per unit length is identical for both wires, satisfying Newton's Third Law.

•

Direction Rule: Parallel currents (flowing in the same direction) attract each other, while anti-parallel currents (flowing in opposite directions) repel each other.

•

Definition of the Ampere: The ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed 1 m1\text{ m} apart in vacuum, would produce between these conductors a force equal to 2×10−72 \times 10^{-7} newtons per metre of length.

📐Formulae

FL=μ0I1I22πd\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}

F=μ0I1I2L2πdF = \frac{\mu_0 I_1 I_2 L}{2\pi d}

μ0=4π×10−7 T m A−1\mu_0 = 4\pi \times 10^{-7}\text{ T m A}^{-1}

1 N/m=1 kg s−21\text{ N/m} = 1\text{ kg s}^{-2}

💡Examples

Problem 1:

Two long parallel wires AA and BB are separated by a distance of 10 cm10\text{ cm} in air. Wire AA carries a current of 5 A5\text{ A} and wire BB carries 8 A8\text{ A} in the same direction. Calculate the magnitude and nature of the force acting on a 20 cm20\text{ cm} section of wire BB.

Solution:

Given: I1=5 AI_1 = 5\text{ A}, I2=8 AI_2 = 8\text{ A}, d=10 cm=0.1 md = 10\text{ cm} = 0.1\text{ m}, and L=20 cm=0.2 mL = 20\text{ cm} = 0.2\text{ m}. Using the formula: F=μ0I1I2L2πdF = \frac{\mu_0 I_1 I_2 L}{2\pi d} F=4π×10−7×5×8×0.22π×0.1F = \frac{4\pi \times 10^{-7} \times 5 \times 8 \times 0.2}{2\pi \times 0.1} F=2×10−7×40×0.20.1F = \frac{2 \times 10^{-7} \times 40 \times 0.2}{0.1} F=2×10−7×80=1.6×10−5 NF = 2 \times 10^{-7} \times 80 = 1.6 \times 10^{-5}\text{ N}

Explanation:

Since the currents flow in the same direction, the force is attractive. The calculation uses the permeability of free space μ0\mu_0 and the standard formula for force between parallel conductors.

Problem 2:

Three long straight parallel wires are arranged. Wire 1 (10 A10\text{ A} upwards) and Wire 3 (5 A5\text{ A} upwards) exert forces on Wire 2 (2 A2\text{ A} downwards) located exactly in the middle. If the force from Wire 1 is 0.0008 N0.0008\text{ N} and the force from Wire 3 is 0.0003 N0.0003\text{ N}, both acting in opposite directions, calculate the net force using vertical subtraction.

Solution:

Force from Wire 1 (F21F_{21}) is repulsive (opposite currents) pushing Wire 2 to the right. Force from Wire 3 (F23F_{23}) is repulsive pushing Wire 2 to the left. Net Force calculation: 0.0008−0.00030.0005\begin{array}{r} 0.0008 \\ -0.0003 \\ \hline 0.0005 \end{array} Net Force = 0.0005 N0.0005\text{ N} to the right.

Explanation:

Because Wire 2 is between Wire 1 and Wire 3, and all currents are vertical, the repulsive forces act in opposite directions along the horizontal axis. We subtract the smaller magnitude from the larger magnitude to find the resultant force.