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Difficulty: MediumMagnetic Force and Electromagnetism

A proton carrying a charge of 1.6×10−19 C1.6 \times 10^{-19}\text{ C} enters a uniform magnetic field of flux density 0.50 T0.50\text{ T} perpendicularly at a speed of 2.0×106 m/s2.0 \times 10^{6}\text{ m/s}. What is the magnitude of the magnetic force acting on the proton?

  1. 1.6×10−13 N1.6 \times 10^{-13}\text{ N}Answer
  2. B
    3.2×10−13 N3.2 \times 10^{-13}\text{ N}
  3. C
    6.4×10−13 N6.4 \times 10^{-13}\text{ N}
  4. D
    1.67×10−21 N1.67 \times 10^{-21}\text{ N}

Answer

The magnetic force acting on the proton is 1.6×10−13 N1.6 \times 10^{-13}\text{ N}.
The magnitude of the force on a charge moving through a magnetic field is given by F=qvBsin⁡θF = qvB\sin\theta. Substituting q=1.6×10−19 Cq = 1.6 \times 10^{-19}\text{ C}, v=2.0×106 m/sv = 2.0 \times 10^6\text{ m/s}, B=0.50 TB = 0.50\text{ T}, and θ=90∘\theta = 90^\circ produces 1.6×10−13 N1.6 \times 10^{-13}\text{ N}.

Step-by-Step Solution

1
Identify the given physical quantities and formula
Charge q=1.6×10−19 Cq = 1.6 \times 10^{-19}\text{ C}, velocity v=2.0×106 m/sv = 2.0 \times 10^{6}\text{ m/s}, field strength B=0.50 TB = 0.50\text{ T}, and angle θ=90∘\theta = 90^\circ. The formula for magnetic force on a moving charge is F=qvBsin⁡θF = qvB\sin\theta.
The magnetic force on a moving charged particle depends on charge magnitude, velocity, magnetic flux density, and the angle between velocity and field vectors.
2
Substitute the values into the force equation
F=(1.6×10−19)×(2.0×106)×0.50×sin⁡(90∘)=1.6×10−13 NF = (1.6 \times 10^{-19}) \times (2.0 \times 10^{6}) \times 0.50 \times \sin(90^\circ) = 1.6 \times 10^{-13}\text{ N}.
Since sin⁡(90∘)=1\sin(90^\circ) = 1, evaluating the product yields the magnetic force in newtons.

Key Concept

Magnetic Force on a Moving Charge
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Topics

Magnetic Force and Electromagnetism

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