Magnetic Force Physics Example 4

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Example 4

medium
An electron (q=1.6×10−19 Cq = 1.6 \times 10^{-19} \text{ C}, m=9.11×10−31 kgm = 9.11 \times 10^{-31} \text{ kg}) moves at 2×107 m/s2 \times 10^7 \text{ m/s} perpendicular to a 0.01 T0.01 \text{ T} field. What is the radius of its circular path?

Solution

  1. 1
    The magnetic force provides centripetal force: qvB=mv2rqvB = \frac{mv^2}{r}.
  2. 2
    Solve for radius: r=mvqB=9.11×10−31×2×1071.6×10−19×0.01r = \frac{mv}{qB} = \frac{9.11 \times 10^{-31} \times 2 \times 10^7}{1.6 \times 10^{-19} \times 0.01}
  3. 3
    r=1.822×10−231.6×10−21=0.0114 m≈1.14 cmr = \frac{1.822 \times 10^{-23}}{1.6 \times 10^{-21}} = 0.0114 \text{ m} \approx 1.14 \text{ cm}

Answer

r≈1.14 cmr \approx 1.14 \text{ cm}
A charged particle in a uniform magnetic field moves in a circle because the magnetic force is always perpendicular to the velocity. The radius depends on mass, speed, charge, and field strength.

About Magnetic Force

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

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