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Magnetic Effects of Electric Current Solved Examples (Class 10 Physics)

Magnetic effects numericals cover magnetic field generation around current-carrying conductors, force on current-carrying conductors in magnetic fields, an

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TL;DR: Magnetic effects numericals cover magnetic field generation around current-carrying conductors, force on current-carrying conductors in magnetic field…

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Magnetic effects numericals cover magnetic field generation around current-carrying conductors, force on current-carrying conductors in magnetic fields, an

Magnetic Effects of Electric Current — Solved Numerical Examples (Step by Step)

Example 1: A current of 5 A flows through a long straight wire. Calculate the magnetic field at a distance of 0.1 m from the wire. (μ₀ = 4π × 10^-7 T·m/A)

Solution: Current I = 5 A, Distance r = 0.1 m. Magnetic field B = (μ₀ × I) / (2π × r) = (4π × 10^-7 × 5) / (2π × 0.1) = (2 × 10^-7 × 5) / 0.1 = 10^-6 / 0.1 = 10^-5 T.

Example 2: A copper wire of length 0.5 m carrying 10 A current is placed perpendicular to a magnetic field of 0.2 T. Calculate the force on the wire.

Solution: Length L = 0.5 m, Current I = 10 A, Magnetic field B = 0.2 T. Force F = B × I × L = 0.2 × 10 × 0.5 = 1 N.

Example 3: A solenoid has 500 turns and carries 2 A current. Find the magnetic field inside (μ₀ = 4π × 10^-7 T·m/A, length = 0.25 m).

Solution: Number of turns N = 500, Current I = 2 A, Length L = 0.25 m. Magnetic field B = (μ₀ × N × I) / L = (4π × 10^-7 × 500 × 2) / 0.25 = (4 × 10^-4 × π) / 0.25 = 1.6π × 10^-3 = 5.03 × 10^-3 T.

Example 4: Two parallel wires carry currents of 5 A each in the same direction, separated by 0.2 m. Calculate the magnetic force per unit length between them. (μ₀ = 4π × 10^-7 T·m/A)

Solution: I₁ = I₂ = 5 A, Distance d = 0.2 m. Force per unit length F/L = (μ₀ × I₁ × I₂) / (2π × d) = (4π × 10^-7 × 5 × 5) / (2π × 0.2) = (2 × 10^-7 × 25) / 0.2 = 5 × 10^-6 / 0.2 = 2.5 × 10^-5 N/m.

Example 5: A rectangular coil with 100 turns and area 0.02 m² is rotated in a magnetic field of 0.5 T. Calculate the maximum EMF induced if it rotates at 50 Hz.

Solution: N = 100, A = 0.02 m², B = 0.5 T, f = 50 Hz. Angular velocity ω = 2πf = 2π × 50 = 100π rad/s. Maximum EMF = N × B × A × ω = 100 × 0.5 × 0.02 × 100π = 100π = 314.16 V.

Tips

  • Right-hand rule: Thumb shows current direction, fingers curl showing magnetic field direction.
  • Parallel currents in same direction attract; opposite directions repel.
  • Force on a current-carrying conductor is maximum when perpendicular to the field.

Frequently Asked Questions

How does an electric motor work?

A current-carrying coil rotates in a magnetic field. The magnetic force on each side of the coil is in opposite directions, creating a turning effect (torque). The commutator reverses current every half rotation to maintain continuous rotation.

Why do compasses point north?

Earth has a magnetic field with its field lines running from south to north. A compass needle aligns with this field, with its north pole pointing toward Earth's north.

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