Question

4.

Pretend you tip the apparatus up at an angle so that the face of the Helmholtz coil is parallel to the direction of the earth's magnetic dip angle. (By "face of the Helmholtz coil" we mean the Area Vector is parallel to Earth's Magnetic Field Vector.) The controls are now pointed up into the air . What would the effect be on the beam of electrons? (Assume you have not changed V or I.)

a) The diameter would decrease.

b) The diameter would increase.

c) The diameter would stay the same.

d) The diameter would be twice as big.

e) The diameter would be half as big.

f) There is not enough information to know the effect.

5.

Given a current of 1.14 A;
a coil radius of 0.18 m;
and N=130,
then what is the strength of the magnetic field in milliTesla?
answer in mT
(Use three sig. figs. or write N/A if there is not enough information.)

6.

Let's say that you are given a magnetic field
B = 0 Tesla.
You also have an
average V / d ² = 24,591 V/m ²;
what is the value for e / m ?
C/kg
(Report your answer with three sig. figs. and use scientific notation "E". ie. 1.98E11)

7. You used a function of current and some other constants to find B.
B, according to this function, was constant. Considering the equation e/m = 8B^-2 * Vd^-2, and considering your data, is B constant for a given current? And how do you know this?

a) B, according to my data, was not a constant value. My Vd^-2 consistently decreased as the diameter increased. e/m can remain constant with a decreasing Vd^-2 only if B decreases as well.

b) B must be constant for a given current because the equation is made only from constants; B=(4/5)^1.5μ_oN*I*r^-1

c) B was not constant, even though it said it was, because diameter of beam is getting bigger so B must change since the bigger d numerically makes the ratio V/d² get smaller.

d) B must be constant because I want it to be constant since the lab manual told me it was constant.

Answer 1