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A proton is moving perpendicular to a uniform magnetic field of 2.5 tesla with 2 MeV kinetic energy. The force on the proton is ______ N. (Mass of proton =1.6×10⁻²⁷ kg, charge of proton =1.6×10⁻¹⁹ C)

Asked in GUJCET 2017 · Lorentz force and work done

• (1) 8×10⁻¹²
• (2) 8×10⁻¹¹
• (3) 3×10⁻¹¹
• (4) 3×10⁻¹⁰

Step-by-step solution

Why the other options are wrong

• (2) 8×10⁻¹¹ follows from a decimal slip in the energy: taking 2 MeV as 3.2×10⁻¹¹ J instead of 3.2×10⁻¹³ J gives v=2×10⁸ m s⁻¹ and F=8×10⁻¹¹ N — a speed faster than light, which is the signal that the conversion went wrong.
• (3) No single slip produces 3×10⁻¹¹ N. Sanity check: K=3.2×10⁻¹³ J fixes v=√2K/m=2×10⁷ m s⁻¹, so F=qvB=8×10⁻¹² N.
• (4) 3×10⁻¹⁰ N is ten times 3×10⁻¹¹ N, and no route reaches either. The energy fixes the speed at 2×10⁷ m s⁻¹, and F=qvB=1.6×10⁻¹⁹×2×10⁷×2.5 leaves no freedom.

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