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A charge of 2 MC moves with a velocity of 3 x 106 m/s in a magnetic field of 0.5 T. Calculate the Lorentz force if the angle between the velocity and magnetic field is 30 degrees.

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Answer

1.5 N\text{1.5 N}

Step-by-step solution

4 steps

  1. Identify the given values and convert units if necessary.

    The charge q=2 μC=2×10−6 Cq = 2\, \mu\text{C} = 2 \times 10^{-6}\, \text{C}.
    The velocity v=3×106 m/sv = 3 \times 10^6\, \text{m/s}.
    The magnetic field B=0.5 TB = 0.5\, \text{T}.
    The angle θ=30∘\theta = 30^\circ.

  2. Recall the formula for the Lorentz force.

    The Lorentz force FF on a charge moving in a magnetic field is given by:

    F=qvBsin⁡θF = qvB \sin\theta

  3. Substitute the values into the formula.

    F=(2×10−6 C)×(3×106 m/s)×(0.5 T)×sin⁡(30∘)F = (2 \times 10^{-6}\, \text{C}) \times (3 \times 10^6\, \text{m/s}) \times (0.5\, \text{T}) \times \sin(30^\circ)

  4. Calculate the value of sin⁡(30∘)\sin(30^\circ) and perform the multiplication.

    We know that sin⁡(30∘)=0.5\sin(30^\circ) = 0.5.

    F=(2×10−6)×(3×106)×(0.5)×(0.5)F = (2 \times 10^{-6}) \times (3 \times 10^6) \times (0.5) \times (0.5) F=(2×3×0.5×0.5)×(10−6×106)F = (2 \times 3 \times 0.5 \times 0.5) \times (10^{-6} \times 10^6) F=(6×0.25)×(100)F = (6 \times 0.25) \times (10^0) F=1.5×1F = 1.5 \times 1 F=1.5 NF = 1.5\, \text{N}

    The Lorentz force is:

    1.5 N\text{1.5 N}
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