Magnetic Field Physics Example 4

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Example 4

hard
A proton (q=1.6×10−19 Cq = 1.6 \times 10^{-19} \text{ C}, m=1.67×10−27 kgm = 1.67 \times 10^{-27} \text{ kg}) enters a uniform magnetic field of 0.2 T0.2 \text{ T} at 90°90° with a speed of 5×106 m/s5 \times 10^6 \text{ m/s}. Calculate: (a) the magnetic force, (b) the radius of its circular path.

Solution

  1. 1
    (a) F=qvBsin⁡θ=1.6×10−19×5×106×0.2×sin⁡90°=1.6×10−13 NF = qvB\sin\theta = 1.6 \times 10^{-19} \times 5 \times 10^6 \times 0.2 \times \sin 90° = 1.6 \times 10^{-13} \text{ N}.
  2. 2
    (b) The magnetic force provides centripetal force: qvB=mv2rqvB = \frac{mv^2}{r}, so r=mvqBr = \frac{mv}{qB}.
  3. 3
    r=1.67×10−27×5×1061.6×10−19×0.2=8.35×10−213.2×10−20≈0.261 mr = \frac{1.67 \times 10^{-27} \times 5 \times 10^6}{1.6 \times 10^{-19} \times 0.2} = \frac{8.35 \times 10^{-21}}{3.2 \times 10^{-20}} \approx 0.261 \text{ m}.

Answer

(a)  F=1.6×10−13 N;(b)  r≈0.26 m(a)\; F = 1.6 \times 10^{-13} \text{ N}; \quad (b)\; r \approx 0.26 \text{ m}
A charged particle moving perpendicular to a uniform magnetic field follows a circular path. The magnetic force is always perpendicular to the velocity (doing no work), so it changes direction but not speed. This principle underlies cyclotrons and mass spectrometers.

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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