Magnetic Force Physics Example 3

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Example 3

medium
A proton (q=1.6×10−19 Cq = 1.6 \times 10^{-19} \text{ C}) moves at 3×106 m/s3 \times 10^6 \text{ m/s} perpendicular to a 0.5 T0.5 \text{ T} magnetic field. Find the magnetic force.

Solution

  1. 1
    Use the magnetic force formula for a moving charge: F=qvBsin⁡θF = qvB\sin\theta. Since the proton moves perpendicular to the field, θ=90°\theta = 90° and sin⁡90°=1\sin 90° = 1.
  2. 2
    Substitute the values: F=1.6×10−19×3×106×0.5F = 1.6 \times 10^{-19} \times 3 \times 10^6 \times 0.5
  3. 3
    Calculate: F=1.6×10−19×1.5×106=2.4×10−13 NF = 1.6 \times 10^{-19} \times 1.5 \times 10^6 = 2.4 \times 10^{-13} \text{ N}

Answer

F=2.4×10−13 NF = 2.4 \times 10^{-13} \text{ N}
The magnetic force on a moving charged particle is perpendicular to both the velocity and the magnetic field (given by the right-hand rule). This force causes the proton to move in a circular path rather than accelerating it in a straight line.

About Magnetic Force

The force exerted on a moving charge or current-carrying conductor by a magnetic field.

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