A metal wire has a mass m = 20 g. The two ends of the wire are separated by L =
ID: 1501185 • Letter: A
Question
A metal wire has a mass m = 20 g. The two ends of the wire are separated by L = 14 cm and are placed in contact with an emf source which is initially switched off. Perpendicular to the wire is a uniform magnetic field of B = 0.20 T. When the emf is switched on, a pulse of charge (q = ?i dt) travels through the wire. Due to the force of the magnetic field on the charge pulse, the wire receives a vertical impulse (J_y = ?F_y dt) which causes the wire to jump and rise to a height of H = 2.5 m before falling back down.
(see attached)
Using the impulse momentum theorem (J = ?p) and conservation of energy (?K + ?U = 0), to determine q the total charge in the pulse that travels through the wire. (Answer 5.0 C)
Explanation / Answer
final potential energy=mgh
let velocity be v
by conservation of energy
kinetic energy=potential energy
0.5mv^2=mgh
v=sqrt(2gh)=7 m/s
so wire moved up with initial velocity of 7 m/s
time taken to complete 1 pulse=0.14/7=0.02 sec
impulse=force*time =qvB*time=change in momentum=mv
q(7*0.2)*0.02=0.02*7
q=5C