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Difficulty: HardMagnetic Force and Electromagnetism

A proton of mass 1.67×10−27 kg1.67 \times 10^{-27}\text{ kg} and charge 1.60×10−19 C1.60 \times 10^{-19}\text{ C} enters a uniform magnetic field of flux density 0.50 T0.50\text{ T} at a speed of 4.00×106 m s−14.00 \times 10^6\text{ m s}^{-1}. If the path of the proton makes an angle of 30∘30^\circ with the magnetic field lines, what is the magnitude of the initial acceleration of the proton?

  1. 9.58×1013 m s−29.58 \times 10^{13}\text{ m s}^{-2}Answer
  2. B
    1.92×1014 m s−21.92 \times 10^{14}\text{ m s}^{-2}
  3. C
    1.66×1014 m s−21.66 \times 10^{14}\text{ m s}^{-2}
  4. D
    9.58×1012 m s−29.58 \times 10^{12}\text{ m s}^{-2}

Answer

The magnitude of the initial acceleration of the proton is 9.58×1013 m s−29.58 \times 10^{13}\text{ m s}^{-2}.
The magnetic force on a moving charged particle is given by F=qvBsin⁡θF = qvB\sin\theta. Substituting q=1.60×10−19 Cq = 1.60 \times 10^{-19}\text{ C}, v=4.00×106 m s−1v = 4.00 \times 10^6\text{ m s}^{-1}, B=0.50 TB = 0.50\text{ T}, and θ=30∘\theta = 30^\circ gives F=1.60×10−13 NF = 1.60 \times 10^{-13}\text{ N}. By Newton's second law, acceleration a=Fm=1.60×10−13 N1.67×10−27 kg=9.58×1013 m s−2a = \frac{F}{m} = \frac{1.60 \times 10^{-13}\text{ N}}{1.67 \times 10^{-27}\text{ kg}} = 9.58 \times 10^{13}\text{ m s}^{-2}.

Step-by-Step Solution

1
Calculate the magnetic force acting on the proton using F=qvBsin⁡θF = qvB\sin\theta.
F=(1.60×10−19 C)×(4.00×106 m s−1)×(0.50 T)×sin⁡30∘=1.60×10−13 NF = (1.60 \times 10^{-19}\text{ C}) \times (4.00 \times 10^6\text{ m s}^{-1}) \times (0.50\text{ T}) \times \sin 30^\circ = 1.60 \times 10^{-13}\text{ N}.
The magnetic force on a moving charge in a magnetic field depends on the component of velocity perpendicular to the field.
2
Apply Newton's second law of motion (F=maF = ma) to find acceleration.
a=Fm=1.60×10−13 N1.67×10−27 kg≈9.58×1013 m s−2a = \frac{F}{m} = \frac{1.60 \times 10^{-13}\text{ N}}{1.67 \times 10^{-27}\text{ kg}} \approx 9.58 \times 10^{13}\text{ m s}^{-2}.
Dividing the magnetic force by the mass yields the magnitude of the resulting acceleration.

Key Concept

Magnetic force on a moving charged particle and Newton's second law
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Topics

Magnetic Force and Electromagnetism

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