Magnetic Field Physics Example 1

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Example 1

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 magnetic field of 0.5 T0.5 \text{ T}. What force does it experience?

Solution

  1. 1
    Force on a moving charge in a magnetic field: F=qvBsin⁡θF = qvB\sin\theta.
  2. 2
    Since the velocity is perpendicular to the field, θ=90°\theta = 90° and sin⁡90°=1\sin 90° = 1.
  3. 3
    F=1.6×10−19×3×106×0.5=2.4×10−13 NF = 1.6 \times 10^{-19} \times 3 \times 10^6 \times 0.5 = 2.4 \times 10^{-13} \text{ N}

Answer

F=2.4×10−13 NF = 2.4 \times 10^{-13} \text{ N}
A magnetic field exerts a force on moving charges perpendicular to both the velocity and the field direction (right-hand rule). This force causes the charge to move in a circular path.

About Magnetic Field

A vector field around magnets and moving charges that exerts force on other moving charges and magnetic materials.

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