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The magnetic force acting on a charged particle of charge −2 μC in a magnetic field of 2T acting in y direction, when the particle velocity is (2î + 3ĵ) × 10⁶ ms⁻¹, is
- 4 N in z direction
- 8 N in y direction
- 8 N in z direction
- 8 N in −z direction
Correct answer: 8 N in −z direction
Solution
The magnetic force is given by **F = q(v × B)**. Here, q = -2 × 10⁻⁶ C, v = (2î + 3ĵ) × 10⁶ m/s, and B = 2ĵ T. The cross product v × B = (2î + 3ĵ) × 2ĵ = -4k̂. Thus, F = -2 × 10⁻⁶ × (-4k̂) = 8 × 10⁻⁶ k̂ N, which is 8 N in the -z direction.
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