PHY2049 - Fall 2016 - HW6 Solutions

PHY2049 - Fall 2016 - HW6 Solutions

Allen Majewski

Department Of Physics, University of Florida

2001 Museum Rd. Gainesville, FL 32611

October 11, 2016

These are solutions to Halliday, Resnick, Walker Chapter 28, No: 10, 16,

30, 35, 38, 39, 51, 65

1

28.10

A proton travels through uniform magnetic and electric fields. The

~ = (?2.50 mT ) i?. At one instant the velocity of

magnetic field is B

the proton is v = (2000 m/s ) j?. At that instant and in unit-vector

notation, what is the net force acting on the proton if the electric

field is (a) (4.00V/m) k? (b) (?4.00 V/m ) k?, and (c) (4.00V/m) i? ?

The Lorentz force invloving both fields is





~

~

~

F = q E + ~v ¡Á B

The proton charge is just e. In all cases ~v = 2000 m/s j? = vy j? and

~

~ = 4.00 k? = Ez k?

B = -2.50 mT i? = Bx i?. For case (a) the electric field is E

?? ? ? ? ? ??

0

0

Bx

~

?

?

?

?

?

?

0 + vy ¡Á 0 ??

F =e

Ez

0

0

You should evaluate the cross product first

1

? ? ? ?

?

?

0

Bx

0

?vy ? ¡Á ? 0 ? = vy Bx (y? ¡Á x?) = vy Bx (?z?) = ? 0 ?

0

0

?vy Bx

Pluggign this into the equations for the Lorentz force we find

?? ? ?

??

0

0

F~ = e ?? 0 ? + ? 0 ??

Ez

?vy Bx

F~ = e (Ez ? vy Bx ) z?





F~ = 1.602x10?19 C 4 V/m ? 2000 m/s -2.5x10?3 T = 1.44x10?18 N k?

~

For (b) and (c), simply evalue this expression for different values of E.

~ is in the x-direction (not z) so you¡¯ll end up with an

Note that in part (c) E

x-component as well as a z-component of the resulting force.

2

28.16

Figure 28-34 shows a metallic block, with its faces parallel to coordinate axes. The block is in a uniform magnetic field of magnitude

0.020 T. One edge length of the block is 25 cm; the block is not

drawn to scale. The block is moved at 3.0 m/s parallel to each

axis, in turn, and the resulting potential difference V that appears

across the block is measured. With the motion parallel to the yaxis, V = 12 mV; with the motion parallel to the z axis, V = 18

mV; with the motion parallel to the x axis, V = 0.What are the

block lengths (a) dx , (b) dy , and (c) dz ?

We make use of the Hall Effect, in which we require that the strength of

force due to the electric field is equal to that of the magnetic field

~ = e ~v ¡Á B

~

eE

|E| = |v| |B|

The hall potential difference is

2

Figure 1: Problem 28.16

V = Ed

Replace E with vB and find

V = vBd

V

=d

vB

0.012 V

= 0.02m

(3.0 m/s) (0.02 T )

0.018 V

dy =

= 0.03m

(3.0 m/s) (0.02 T )

dz =

3

Then dx must be 0.025 m by process of elimination.

3

28.30

Figure 2: Problem 30

In Fig. 28-39, an electron with an initial kinetic energy of 4.0

keV enters region 1 at time t = 0. That region contains a uniform

magnetic field directed into the page, with magnitude 0.010 T.

The electron goes through a half-circle and then exits region 1,

headed toward region 2 across a gap of 25.0 cm. There is an electric

potential difference ?V = 2000 V across the gap, with a polarity

such that the electrons speed increases uniformly as it traverses

the gap. Region 2 contains a uniform magnetic field directed out

of the page, with magnitude 0.020 T. The electron goes through a

half-circle and then leaves region 2. At what time t does it leave?

4

If the electron spends an amount of time t1 in region 1, tgap in the gap,

and t2 in region two, the total time of this journey is

ttotal = t1 + tgap + t2

In regions 1 and 2, the electron makes a half circle

¦ÐR

d

=

v

v

In circular motion due to a magnetic field, we have

t=

v2

= qvB

R

so v = qBR/m and the t becomes

m

t=

¦ÐR

¦Ðm

d

=

=

v

qBR/m

qB

We can use this to get t1 and t2 :

t1 =

¦Ðme

¦Ð 9.109e-31 kg

=

= 1.786 ns

eB1

1.602e-19 C ¡Á 0.01 T

¦Ðme

¦Ð 9.109e-31 kg

=

= 0.893 ns

eB2

1.602e-19 C ¡Á 0.02 T

The time in the gap requires us to do a little work. The electron accelerates trough a potential difference of 2 kV. So it gaines 2 keV of energy. It

enters the gap with Ei = 4 keV and thus leaves with Ef = 6 keV. It is useful

here to know 1 keV = 1.602e-16 J. We express speed in terms of E using

v = 21 mv 2 and find

r

2Ei

vi =

= 3.7509e7 m/s

me

r

2Ef

vf =

= 4.5939e7 m/s

me

We can use the kinematic equation

t2 =

1

(vf + vi ) tgap = d

2

5

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