Question Bank NAT • 1 Mark Section 3: Electromagnetic Theory Electrostatics and Magnetostatics

QB_PH_084: Electrostatics and Magnetostatics

Question Statement

A proton enters magnetic field $B = 0.5\text{ T}$ at velocity $v = 10^6\text{ m/s}$ perpendicularly. The force in pN is:
Correct Answer
0.08, 0.08
Verified Key

Step-by-Step KaTeX Derivation & Detailed Explanation

### Solution: $F = qvB = (1.6 \times 10^{-19})(10^6)(0.5) = 0.8 \times 10^{-13}\text{ N} = 0.08\text{ pN}$.
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