A conducting rod PQ of length L=1.0 m moves with speed v=20 m/s in a uniform magnetic field B=4.0 T directed into the page. A capacitor C=10 µF is connected as shown. Then:

Question

G1 Physics · Accepted Answer

The motional emf is ε=BLv=(4)(1)(20)=80 V. The capacitor charges until its voltage is 80 V, so q=Cε=(10 µF)(80 V)=800 µC. The polarity from v⃗×B⃗ makes qₐ positive and qᵦ negative.