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Difficulty: MediumMagnetic Force and Electromagnetism

A particle carrying a positive charge of 3.2×10−19 C3.2 \times 10^{-19}\text{ C} moves with a velocity of 4.0×106 m/s4.0 \times 10^6\text{ m/s} into a uniform magnetic field of magnetic flux density 0.50 T0.50\text{ T}. If the velocity vector of the particle makes an angle of 30∘30^\circ with the direction of the magnetic field, what is the magnitude of the magnetic force exerted on the particle?

  1. 3.2×10−13 N3.2 \times 10^{-13}\text{ N}Answer
  2. B
    6.4×10−13 N6.4 \times 10^{-13}\text{ N}
  3. C
    5.5×10−13 N5.5 \times 10^{-13}\text{ N}
  4. D
    1.6×10−13 N1.6 \times 10^{-13}\text{ N}

Answer

The magnitude of the magnetic force exerted on the particle is 3.2×10−13 N3.2 \times 10^{-13}\text{ N}.
The magnetic force on a charged particle moving through a magnetic field is given by F=qvBsin⁡θF = q v B \sin\theta. Substituting q=3.2×10−19 Cq = 3.2 \times 10^{-19}\text{ C}, v=4.0×106 m/sv = 4.0 \times 10^6\text{ m/s}, B=0.50 TB = 0.50\text{ T}, and sin⁡(30∘)=0.50\sin(30^\circ) = 0.50 yields F=3.2×10−13 NF = 3.2 \times 10^{-13}\text{ N}.

Step-by-Step Solution

1
Identify the given physical quantities
q=3.2×10−19 Cq = 3.2 \times 10^{-19}\text{ C}, v=4.0×106 m/sv = 4.0 \times 10^6\text{ m/s}, B=0.50 TB = 0.50\text{ T}, θ=30∘\theta = 30^\circ
Extract values needed for the magnetic force formula.
2
Apply the magnetic force formula for a moving charge
F=qvBsin⁡θF = q v B \sin\theta
The force experienced by a moving point charge in a uniform magnetic field depends on charge, speed, field strength, and the angle between velocity and field.
3
Substitute the values and calculate
F=(3.2×10−19)×(4.0×106)×0.50×sin⁡(30∘)=(6.4×10−13)×0.50=3.2×10−13 NF = (3.2 \times 10^{-19}) \times (4.0 \times 10^6) \times 0.50 \times \sin(30^\circ) = (6.4 \times 10^{-13}) \times 0.50 = 3.2 \times 10^{-13}\text{ N}
Since sin⁡(30∘)=0.5\sin(30^\circ) = 0.5, evaluating the expression yields 3.2×10−13 N3.2 \times 10^{-13}\text{ N}.

Key Concept

Magnetic Force on a Moving Charge (F=qvBsin⁡θF = q v B \sin\theta)
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Topics

Magnetic Force and Electromagnetism

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