(^278) Aptitude Test Problems in Physics
through the current source. Obviously, el = C.
From the energy conservation law, we obtain
Q -1-4 2
1Fq
2C '
where Q is the amount of heat liberated in both
resistors. Since they are connected in parallel,
Q 1 /Q 2 = R 2 /R 1 , whence
Qa= C
U'; R1
Rs
CUR'
2 (Ri+ R 2 )3.48. During the motion of the jumper, the mag-
netic flux through the circuit formed by the jump-
er, rails, and the resistor changes. An emf is
induced in the circuit, and a current is generated.
As a result of the action of the magnetic field on
the current in the jumper, the latter will be de-
celerated.
Let us determine the decelerating force F.
Let the velocity of the jumper at a certain instant
be v. During a short time interval At, the jumper
is shifted along the rails by a small distance Ax
v At. The change in the area embraced by the
circuit is vd At, and the magnetic flux varies by
= Bvd At during this time. The emf induced
in the circuit is
V=-- At = —Bvd.
According to Ohm's law, the current through the
jumper is I = VB. The force exerted by the mag-
netic field on the jumper is
B 2 d 2 v
F= IBd=
R •According to Lenz's law, the force F is directed
against the velocity v of the jumper.