A proton (charge 1.6 × 10−19 C, mass 1.67 × 10−27 kg) moves at 2 × 106 m s−1 at right angles to a field of 0.5 T. Find the force on it and the radius of its path.
- F = qvB = 1.6 × 10−19 × 2 × 106 × 0.5 = 1.6 × 10−13 N.
- This force is the centripetal force: qvB = mv2/r, so r = mv/(qB).
- mv = 1.67 × 10−27 × 2 × 106 = 3.34 × 10−21 kg m s−1; qB = 1.6 × 10−19 × 0.5 = 8.0 × 10−20 C T.
- r = (3.34 × 10−21) ÷ (8.0 × 10−20) = 0.042 m.
Answer: F = 1.6 × 10−13 N; r ≈ 4.2 cm.