Paper 2 Magnetic Force Answers

Q1 Force on a Wire

1. $F=BIL=(0.42)(3.5)(0.080)=0.118\mathrm{N}$. [2]
2. It becomes zero because $\theta =0$ and $\sin \theta =0$. [1]
3. Fleming left-hand rule, or vector product direction for $\vec{I}\times \vec{B}$. [1]

Q2 Charged Particle in a Magnetic Field

1. $F=Bqv=(0.030)(1.60\times 10^{-19})(4.0\times 10^6)=1.92\times 10^{-14}\mathrm{N}$. [2]
2. $r=mv/(Bq)=9.11\times 10^{-31}(4.0\times 10^6)/(0.030\times 1.60\times 10^{-19})=7.59\times 10^{-4}\mathrm{m}$. [3]
3. Magnetic force is perpendicular to velocity, so it changes direction of velocity but does no work. [2]

Q3 Velocity Selector

1. Undeflected condition: $qE=qvB$, so $v=E/B=6.0\times 10^4/0.20=3.0\times 10^5\mathrm{m}{\mathrm{s}}^{-1}$. [2]
2. If $v$ differs, magnetic force $qvB$ no longer balances electric force $qE$, giving a resultant transverse force. [2]