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    Question

    A particle with charge q = 2 μC moves through a

    magnetic field B = 0.5 T with velocity v = 4 m/s at angle 60°. Find magnitude of magnetic force.
    A 2.7 μN Correct Answer Incorrect Answer
    B 3.46 μN Correct Answer Incorrect Answer
    C 4.89 μN Correct Answer Incorrect Answer
    D 6.31 μN Correct Answer Incorrect Answer

    Solution

    Given:

    • Charge: q = 2 μC = 2 × 10⁻⁶ C
    • Magnetic field: B = 0.5 T
    • Velocity: v = 4 m/s
    • Angle between v and B: θ = 60°
    Formula for magnetic force: The magnitude of the magnetic force on a moving charged particle is: F = qvB sin θ Calculation: F = qvB sin θ F = (2 × 10⁻⁶)(4)(0.5) sin(60°) F = (2 × 10⁻⁶)(4)(0.5)(√3/2) F = (2 × 10⁻⁶)(4)(0.5)(0.866) F = (2 × 10⁻⁶)(1.732) F = 3.464 × 10⁻⁶ N Converting to μN: Since 1 μN = 10⁻⁶ N: F = 3.464 × 10⁻⁶ N = 3.464 μN

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