JEE Main 2024 Physics question
A charge of 4.0 μC is moving with a velocity of 4.0 × 10⁶ ms⁻¹ along the positive y-axis under a magnetic field B of strength (2k̂) T. The force acting on the charge is x î N. The value of x is .
Answer: 32
Solution
F = |q(v⃗ × B⃗)| = 4 × 10⁻⁶ × 4 × 10⁶ × 2 = 32 N (along î).