## PHYS208 Fundamentals of Physics II

### Hint for Ch. 29, 12P

Set the Lorentz force on the electron to zero.
This yields the velocity selector condition.
Find the velocity from the kinetic energy of the electron (100 eV).

### Hint for Ch. 29, 16P

a) Use the velocity selector condition; the transverse electric field is
1.0 mV/m.

b) Solve Eq. 29-11 to find *n* = *i*/(*A e v*_{d}),
where *A* is the cross-sectional area of the strip.

### Hint for Ch. 29, 19E

a) The electron speed can be found from the kinetic energy after acceleration;
the change in kinetic energy is equal to the difference in potential energy between
the two positions. Don't forget to convert *eV* to joules.

b) Use the relation between centripetal acceleration and the
radius of the path in uniform circular motion.

### Hint for Ch. 29, 21E

a) Use the relation between centripetal acceleration and the
radius of the path in uniform circular motion.
The speed of light is 3 x 10^{8} m/s.

b)
Kinetic energy in joules can be converted to *eV* by the conversion
factor 1 *eV* = 1.60 10^{-19} J.

"http://www.physics.udel.edu/~watson/phys208/exercises/hints/0330.html"

Last updated March 30, 1998.

Copyright George Watson, Univ. of Delaware, 1997.