An electron enters a uniform magnetic field B=0.01 T perpendicular to its velocity v=2×107 m/s. Then a second electron enters the same field at 30° to B with the same speed. For each, find (a) the radius of the circular path (or helix), (b) the period of circular motion, (c) the pitch (for the second case). Take me=9.1×10−31 kg, e=1.6×10−19 C.
Solution — Step by Step
Magnetic force provides centripetal: evB=mv2/r, so r=mv/(eB).
r1=1.6×10−19×0.019.1×10−31×2×107
=1.6×10−211.82×10−23=0.01138 m≈1.14 cm.
T=2πm/(eB) — independent of velocity.
T=1.6×10−19×0.012π×9.1×10−31=1.6×10−215.72×10−30≈3.57×10−9 s=3.57 ns.
Only v⊥ matters for radius: r2=mv⊥/(eB)=r1×sin30°=0.5×1.14=0.57 cm.
Pitch is the distance traveled along B per revolution: p=v∥×T.
p=1.732×107×3.57×10−9≈0.0618 m=6.18 cm.
Final answers: Electron 1: r=1.14 cm, T=3.57 ns. Electron 2: r=0.57 cm, T=3.57 ns, pitch =6.18 cm.
Why This Works
The trap with helical motion is that T does not change with the angle. The parallel velocity component doesn’t experience any magnetic force (v∥×B=0), so it’s pure inertial motion along B. The perpendicular component does the circling, and its period depends only on m, q, and B.
So when an electron enters at any angle (other than 0°), the result is a helix of radius mvsinθ/(qB) and pitch vcosθ⋅T.
Alternative Method
We can use the cyclotron frequency ω=eB/m directly. ω=(1.6×10−19×0.01)/(9.1×10−31)≈1.76×109 rad/s, giving T=2π/ω≈3.57 ns. Faster than computing r first and dividing.
Common Mistake
Students often write r=mv/(eB) for helical motion using the fullv. That’s wrong — only v⊥ goes into the radius formula. Also, many forget that the period T stays the same regardless of the entry angle, so they recompute it for each angle (wasted effort).
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