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up** - by Brian Hupy, 2002 and The Smart
German (Maike) and Trina KALILIMOKU, 2004**

^{}**1. (a) What is the force per meter on a wire
carrying a 9.80-A current when perpendicular to a .80-T magnetic field?**

Since the question asks for the force per meter, the equation (F =

Ilx B, since the wire is perpendicular to the field) needs to be modified.F /

l=Ix B= 9.80 x .80

= 7.84 N/m

(b) What if the angle between the wire and the field is 45?Since the wire is not perpendicular to the field in this question, we need to modify the other equation.

F /

l=IB sin (q)= 9.80 x .80 x sin (45)

= 5.54 N/m

**3. How much current is flowing in a wire 4.20 m long if the maximum
force on it is .900 N when placed in a uniform .0800-T field?**

The maximum force will occur when the wire is placed perpendicular to the field, so the formula we use is F =

Ilx B.Except now we're looking for current, so we need to modify the equation.

I= F/(lx B)= .900 / (4.20 x .0800)

= 2.69 A

**4. The force on a wire carrying 25.0A is a maximum of 4.14
N when placed between the pole faces of a magnet. If the pole faces are 22.0 cm
in diameter, what is the approximate strength of the magnetic
field? **

The pole faces are the area of the magnetic field, so the largest length that can affect the wire is the diameter, 22.0 cm. Again we use the formula F =

Ilx B, modified.B = F / (

Il)= 4.14 / (25.0 x .220)

= .75 T

(Table of contents)

**7. Find the direction of the force on a negative charge for
each diagram shown in shown in Fig. 20-50, where v is the velocity of the charge
and B is the direction of the magnetic field.**

Use the right hand rule. Point your index finger in the direction of v and your middle finger in the direction of B. Your thumb will point in the opposite direction of the answer because the charge is negative. For those of you who are lazy, they are listed below.

Left.

Left.

Up.

Into the Page.

Out of the Page.

Down.

**8. Determine the direction of B for each case in Fig.
20-51, where F represents the force on a positively charged particle moving with
velocity v.**

Once again, use the right hand rule, Thumb is F, index finger is v. Answer is middle finger.

Down.

Into the Page.

Right.

**9. Alpha particles of charge q = +2e and mass m = 6.6 E-27
kg are emitted from a radioactive source at a speed of 1.6 E7 m/s. What magnetic
field strength would be required to bend these these in a circular path of
radius r = .25 m?**

Set the force on the particle due to the magnetic field equal to the centrifu... uh centripetal force necessary to keep the particle moving in a circle.

qvb = mv

^{2}/rcancel the v's where possible

qb = mv/r

b = (mv)/(qr)

= (6.6 E-27 * 1.6 E7)/(2e * .25)

= 1.32 T

(Table of contents)

**13. A proton moves in a circular path perpendicular to a
1.15T magnetic field. The radius of its path is 8.4 mm. Calculate the energy of
the proton in eV.**

Use the same set-up from problem 9.

qb = mv/r

v = qbr/m

= e * 1.15 * .0084 / 1.6726 E-27 kg

= 925315 m/s

Now find its kinetic energy and convert to eV

E = 1/2 mv^2

= 7.16 E-16 J

=4469 eV

(Table of contents)

**15. A particle of charge q moves in a circle of radius r
in a uniform magnetic field B. Show that its momentum is p=qBr.**

Once again, use the same set-up as problem 9.

mv^2/r = qvb

mv/r = qb

mv = qbr

p = qBr

(Table of contents)

**50. Protons move in a circle of radius 5.10 cm in a 0.566
T magnetic field. What value of electric field could make their paths straight?
In what direction must it point?**

When the protons are moving in a circle, the Force the magnetic field is exerting (F = qvB) provides the centripetal force for it to go in a circle. F = ma, a = v

^{2}/r, so F = mv^{2}/r. If you put these together, you get qvB = mv^{2}/rIf you solve this for the velocity of the proton you get

v = rqB/m

Now, plug in the values given in the problem.

v=.0510(1.602E-19)(.566)/1.6726E-27

So, v= 2764757 m/s

If the electric field makes it go straight, then it must exert an equal but opposite force, E = F/q, so F = Eq, and F = qvB, so Eq = qvB, and E = vB

E=vB plug in the values

E= 2764757(.566)=1.56E6 V/m

As far as the direction, if the protons are initially going to the right, and if the B field is out of the page (at us) then the magnetic force would be down using the right hand rule. The electric field would be up the page then to exert an upward force on the protons.

(Table of contents)

**53. (a) What value of magnetic field would make a beam of electrons,
traveling to the right at a speed of 4.8E6 m/s, go undeflected through a
region where there is a uniform electric field of 10,000 V/m pointing
vertically up? (b) What is the direction of the magnetic field if it is known
to be perpendicular to the electric field? (c) What is the frequency of the
circular orbit of the electrons if the electric field is turned off?**

(a) The formula we'll be using is qE=qvB since the electric field and the magnetic field balance.

10,000V/m=(4.8E6)B.......(charges cancel!!) which gives B=2.1E-3T

(b) Because the electric field is up, the electrons would be deflected down by the electric field. The magnetic field must then exert an upward force, and the magnetic field must therefore be out of the page to do this (remember, it is all backwards for an electron)

(c) When the protons are moving in a circle, the Force the magnetic field is exerting (F = qvB) provides the centripetal force for it to go in a circle. F = ma, a = v

^{2}/r, so F = mv^{2}/r. If you put these together, you get qvB = mv^{2}/rIf you solve this for the velocity of the proton you get

v = rqB/m

The time to complete a circle is

T=2πr/v= 2πm/qB, so the frequency is f=1/T=qB/2πm=(1.60E-19)(2.08E-3)/2π(9.11E-31)

and the answer is..........................................................................................5.8x10

^{7}Hz

**17. A sort of "projectile launcher" is shown in Fig. 20-52. A
large current moves in a closed loop composed of fixed rails, a power supply,
and a very light, almost frictionless bar touching the rails. A magnetic field
is perpendicular to the plane of the circuit. If the bar has a length of 20
cm, a mass of 1.5 g, and is placed in a field of 1.7 T, what constant current
flow is needed in order for it to accelerate to 30 m/s in a distance of 1.0 m?
In what direction must the field point?**

<Pic. of Fig. 20-52 goes here!!>

Yeah SUVAT!! So first fill out a SUVAT and solve for A

S 1

U 0

V 30

A ?

T don't care!

Use the formula v

^{2}= u^{2}+ 2as and plug in the values! So we find out A=450 m/s² now we find the force! So we have the mass and we have the acceleration and hey, we have the formula F=ma! wow, plug the values in and solve for F....which equals .675 N. But the question asks for the current.....so use the formula F=IlB and solve for the current! The answer is I=2.0 A, down.

**19.Jumper cables used to start a stalled vehicle often carry a 15-A
current. How strong is the magnetic field 15cm away? What percentage of the
Earth's magnetic field is this?**

Use the formula B = m

_{0}I/(2pr)and solve for B=(4pE-7)(15)/((2p)(.15))=2E-5 T.Now to find the percentage...Using the magnetic field we just found, divide that by the earth's magnetic field(.55E-4 T) 2E-5/.55E-4=.36 or 36%.

**20. If a magnetic field no larger than that of the Earth (0.55E-4 T) is to
be allowed 30 cm from an electrical wire, what is the maximum current the wire
can carry?**

Use the formula B = m

_{0}I/(2pr)and solve for I...I=2prB/m_{0}

_{Plug in the values and solve!! I=82.5 A, or in sig figs 83 A.}

**21. What is the magnitude and direction of the force between two parallel
wires 45 m long and 6.0 cm apart, each carrying 35A in the same direction?**

Hmm...2 parallel wires....must be the formula....F/l = (m

_{0}/2p)(I_{1}I_{2}/r), only we need to find the Force so just multiply both sides by the length, l.F=m

_{o }I(1)I(2)l/2pr and solve for F. F=m_{o }(35)(35)(45)/2p(.6)=.18N, attraction

**22. A vertical straight wire carrying an upward 12-A current exerts an
attractive force per unit length of 8.8E-4 N/m on a second parallel wire 7.0
cm away. What current (magnitude and direction) flows in the second wire?**

OK, use the formula F/l = (m

_{0}/2p)(I_{1}I_{2}/r)and solve for I(2) and then just plug in the values!!I

_{2}=2prF/m_{0}Il=25.666 or 26 A

**27. A stream of protons passes a given point in space at a rate of
10 ^{9}
protons/s. What magnetic field do they produce 2.0 m from the beam? **

I=10

^{9}(1.602E-19)=1.602E-10 AThen plug this into the formula to find the magnetic field. B=(2E-7)(1.602E-10)/2.0, B=1.602E-17 T

**36. A 30.0-cm long solenoid 1.25 cm in a diameter is to produce a field of
0.385 T at its center. How much current should the solenoid carry it is has
1000 turns of the wire?**

The formula used here will be....B = m

_{0}(NI/l) only we don't need to solve for B, so solve this formula for I (current)...so, I=Bl/m_{0}N and plug in the values....I=.385(.3)/m_{0}(1000)and solve!! I=91.9 A

**24. What is the acceleration (in g's) of a 175-g model airplane charged to
18.0C and traveling at 1.8 m/s as sit passes within 8.6 cm of a wire, nearly
parallel to its path, carrying a 30-A current?**

First use the formula B = m

_{0}I/(2pr)and plug in all of the values. B=2E-7(30)/.086m and solve for B. Since F = qvB and F=ma and a = v^{2}/r, just set the first equation equal to the second and plug in v^{2}/r for a. We need to find the acceleration so solve for a.qvB= F = ma

(18)(1.8)(6.97E-5) = 1.75a, a=.00129, a=1.3E-3 g's

**29. A long pair of wires serves to conduct 25.0 A of dc current to (and
from) an instrument. If the wires are of negligible diameter but are 2.0mm
apart, what is the magnetic field 10.0 cm from their midpoint, in their plane?
Compare to the magnetic field of the Earth.**

Use the formula B = m

_{0}I/(2pr) twice, both times with a current of 25.0 A, but one wire is 9.9 cm from the point, the other 10.1 cm from the point.B=B

_{1}-B_{2}because the currents are in opposite directions, so the fields will be in opposite directions. And since we need to find the midpoint, subtract them! Use the formula B = m_{0}I/(2pr)and plug in the values!!B=[4pE-7(25)/(2p(.0101))] - [4pE-7(25)/(2p(..0099))]=1.0E-6T up

This is (1.0E-6)/(5.0E-5)=.02=2% of the Earth's field

**31. Three long parallel wires are 38.0 cm from one another. (Looking along
them, they are at three corners of an equilateral triangle.) the current in
each wire is 8.00 A, but that in wire A is opposite to that in wire B and C.
Determine the magnitude force per unit length on each wire due to the other
two.**

Diagram goes hereUse the formula F/l = (m

_{0}/2p)(I_{1}I_{2}/r)and plug in the given values and solve for F.l=1m and both currents (I) are equal to 8 Amps and the radius is .38 meters... there you go, now solve...

F= 3.37E-5 N

Now add the vectors! Use the formula F

_{a}/L=2(F/L)cos30 and the force we found above! F_{a}/L=2(3.37E-5)cos30=5.8E-5, upF

_{b}/L=(F/L)=3.4E-5 N/m, 60 degrees below the line toward CF

_{c}/L=(F/L)=3.4E-5 N/m, 60 degrees below the line toward B

**33. Two long wires are oriented so that they are perpendicular to each
other, and at their closest, they are 20.0 cm apart. What is the magnitude of
the magnetic field at a point midway between them if the top one carries a
current of 20.0 A and the bottom one carries 5.0 A?**

diagram goes here

Use the formula B = m

_{0}I/(2pr)twice since there are two given currents..

First....B

_{b}=(4pE-7)(5.0)/(2p(.1))=1E-5 T and same formula with 20 Amps instead of 5 Amps to solve for the direction of the field for each wire. So B_{t}= (4pE-7)(20.0)/(2p(.1))=4.00E-5 T. Now since the magnetic fields are perpendicular, we use VECTORS to find the magnitude! In the PHYSICS BOOK WITH ALL THE ANSWERS, they give us the formula B=( B_{t}²+B_{b}²)^(1/2), plug in the answers from above and the answer is ....4.1E-5 T

**61. Calculate the force of an airplane which has acquired a net charge of
155 C and moves with a speed of 120 m/s perpendicular to the Earth's magnetic
field of 5.0E-5 T.**

Use the formula qvB=F where q=155 C, v=120m/s and B=5.0E-5 T ... plug everything in

F=155(120)(5.0E-5)=.93 N