Bukhovtsev-et-al-Problems-in-Elementary-Physics

352 ANSWERS AND SOLUTIONS
the ring is arranged along the force lines of the field (the induced current is
zero) or when the plane of the ring is strictly perpendicular to the force lines
(the moment of the forces is zero).
According to Lens's law, the first position of the ring will be stable in an
increasing magnetic field, and the second unstable.
In a decreasing magnetic field, on the contrary, equilibrium will be stable
when the angle between the plane of the ring and. the force lines is a right
one, and unsta ble when the plane of the ring is para lIel to the force lines.
583. Let the velocity of the conductor be v at a certain moment of time.
The e. m. f. (in volts) at the same moment of time will thus be cC= 10- 8 Hlo,
and the current I=ir x 10- 8 Hlo,
The action of the magnetic field on the
conductor carrying a current will induce a force f that prevents free dropping
of the conductor:
f= 10-' H~2fJ
being considered, the acceleration can be de­
Hence, at the moment of time
termined from the relationship
H 2 l 2U
ma=mg-f=mg-IO-8~
It is easy to see that as the velocity increases, the acceleration a will dlminish
and become zero at the moment of equality of the forces f=mg. From
this moment on, the conductor will move with a constant velocity Vk equal to
mgRXI0 9
Vk H 2l 2
584. "The e. m. f. of induction appearing in the conductor (measured in volts)
is It= 10- 8 Hlu.
The charge on the plates of the capacitor can be found from the relationship
Q=l1C= 10- 8 HlvC
The current flowing in the circuit is
1=~~=10-8 HlC ~~=10-8HICa
where a Is the sought acceleration.
The interaction of this current with the magnetic field will produce a force F l
acting on the moving conductor. On the basis of Lenz's law, this force will
be directed oppositely tc the force F.
The force F I =kJHI = lo-eHt/laC, if C is measured in farads. The sought
acceleration can be found from the equation ma=F-Fl.
Hence,
F
is a constant quantity.