JEE Main 2024 Physics question
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.
Answer: 35
Solution
The conductor makes 45° with the field direction (north). F/ℓ = BI sin45° = 3.5 × 10⁻⁵ × √2 × (1/√2) = 3.5 × 10⁻⁵ N/m = 35 × 10⁻⁶ N/m.