The horizontal component of earth's magnetic field at a place is 3.5 × 10⁻⁵ T. A very long straight conductor carrying current of √2 A in the direction from South east to North West is placed. The force per unit length experienced by the conductor is __________ × 10⁻⁶ N / m.

Solution

Physics sir · 3 steps

Find the angle between the wire and north; the √2 in the current is there to cancel sin 45°.

1. Direction of field. The horizontal component of the earth's field points from geographic south to north. Only this component is given, so it is the field we use.
2. Find the angle. The wire runs from south-east to north-west, which is exactly halfway between north and west. So the wire makes an angle θ = 45° with the north-pointing field.

   F/L = I B sin θ = √2 × 3.5 × 10⁻⁵ × 1/√2

3. So. The √2 cancels, leaving F/L = 3.5 × 10⁻⁵ N m⁻¹ = 35 × 10⁻⁶ N m⁻¹. The answer is 35.

The common slip: 49.5. Taking the wire as perpendicular to the field (sin 90° = 1) gives √2 × 35 ≈ 49.5, not 35.

The idea to keep. Force on a wire depends on the angle between the current and the field, F = IlB sin θ. Turn compass directions into an angle first. The earth's horizontal field always points north.

Answer from the NTA final answer key. Solution by PrepWiser.