Homework ## Chapter

20:Magnetism

Ch 20: 4, 12, 21, 24, 30, 64

20.4 A charge of 1.5 x 10^{ - 10}C moves at 1.0 km/s through a magnetic field in such a direction that the magnetic force on it is a maximum. What is the magnetic field strength if the maximum force on the charge is 8 x 10^{ - 8}N?F = q v B sin F

_{max}= q v BB = F

_{max}/[q v]B = 8 x 10

^{ - 8}N / [(1.5 x 10^{ - 10}C)(1 000 m/s)]B = 0.533 T

20.12 An electron moves in a circular path, perpendicular to a magnetic field of B = 1.25 T, with a a speed of 1.2 x 10^{4}m/s. What is the radius of the electron's circular path?F _{mag}= q v B sinSince = 90, sin = 1

F _{mag}= q v BThis is also the centripetal force,

F _{c}= m v^{2}/ rq v B = m v

^{2}/ rr = m v / [q B]

r = (9.11x10

^{-31}kg) (1.2x10^{4}m/s) / [(1.6x10^{-19}C) (1.25 T)]r = 5.47 x 10

^{-8}m

20.21 A 100 T magnetic field is confined to a 5.0-cm region as shown in Figure 20.47. An electron passes through the region with a speed of 5.16 x 10^{6}m/s as shown. Determine the angular deflection .

[ The 100 T magnetic field is so very strong that it does not just deflect the electron a little like my analysis assumes. Rather, the electron would make a circle -- or half a circle. So the numbers are unrealistic! Try two variations of this:

20.21 a; An electron enters a region where the magnetic field is 100 T, moving perpendicular to the magnetic field with a speed of 5.16 x 10^{6}m/s. What is the radius of the circle that it now moves it? It will make a half circle "orbit" and then exit the region with the magnetic field. What is the radius of that semicircular path.

20.21 b; Earth's magnetic field is about 5 x 10^{ - 5}T or 50 x 10^{ - 6}T. A somewhat larger magnetic field of 100 x 10^{ - 6}T is confined to a 5.0-cm region as shown in Figure 20.47. An electron passes through the region with a speed of 5.16 x 10^{6}m/s as shown. Determine the angular deflection .

This second question is the one whose solution I had initially.]

20.21 a; An electron enters a region where the magnetic field is 100 T, moving perpendicular to the magnetic field with a speed of 5.16 x 10^{6}m/s. What is the radius of the circle that it now moves it? It will make a half circle "orbit" and then exit the region with the magnetic field. What is the radius of that semicircular path.With the magnetic field this

strong, the electron is not going to be deflected just a little. It is going to move in a circle -- actually in a semi-circle. The magnetic force isF

_{mag}= q v BF

_{mag}= (1.6 x 10^{ - 19}C)(5.16 x 10^{6}m/s)(100 T)F

_{mag}= 8.26 x 10^{ - 11}NThis force provides the

centripetalforce,F

_{c}= m v^{2}/rr = m v

^{2}/F_{c}r = (9.11 x 10

^{ - 31})(5.16 x 10^{6}m/s)^{2}/ (8.26 x 10^{ - 11}N)r = 2.9 x 10

^{ - 7}mThat's a

verysmall radius -- but the mass of the electron isverysmall, too

20.21 b; Earth's magnetic field is about 5 x 10^{ - 5}T or 50 x 10^{ - 6}T. A somewhat larger magnetic field of 100 x 10^{ - 6}T is confined to a 5.0-cm region as shown in Figure 20.47. An electron passes through the region with a speed of 5.16 x 10^{6}m/s as shown. Determine the angular deflection .

This second question is the one whose solution I had initially.]

Figure 20.47 Problem 20.21

20.21bassumes that v_{x}remains constant. Then we can use that to determine thetimethe electron spends with this magnetic force on it. That will let us calculate v_{y}and, from that, the deflection angle.First, find the force on the electron,

F = q v B sin Since = 90, sin = 1.0

F = (1.6 x 10 ^{ - 19}C) (5.16 x 10^{6}m/s) (100 x 10^{ - 6}T) (1.0)F = 8.256 x 10

^{ - 17}NWhile this seems like a very small force, remember it is acting on a

verysmall mass, the mass of the electron,F = m a a = F / m

a = 8.256 x 10

^{ - 17}N / 9.11 x 10^{ - 31}kga = 9.16 x 10

^{13}m/s^{2}How long does this acceleration act?

v = s/t t = s/v

t = 0.05 m / ( 5.16 x 10

^{6}m/s )t = 9.69 x 10

^{ - 9}sIf that acceleration is all in the y-direction, what will be the final velocity component in the y-direction?

v _{y}= v_{yo}+ a_{y}tv

_{y}= 0 + (9.16 x 10^{13}m/s^{2}) (9.69 x 10^{-9}s)v

_{y}= 8.88 x 10^{5}m/sThe deflection angle is given by

tan = v _{y}/ v_{x}If we use the approximation that v

_{x}= v_{xo}= 5.16 x 10^{6}m/s, thentan = (8.88 x 10 ^{5}m/s) / (5.16 x 10^{6}m/s)tan = 0.172

= 9.76

^{o}With such a small deflection angle, our two approximations are, indeed, reasonable. The acceleration is almost entirely in the y-direction and the x-component of the velocity remains nearly constant.

20.24 In a region where Earth's magnetic field is 5.0 x 10^{ - 5}T and makes an angle of 60° with Earth's surface, determine the force on a straight 2.0 m wire that carries a current of 8.0 A straight up.F = I l B sin

F = (8.0 A) (2.0 m)( 5.0 x 10

^{ - 5}T) (sin 30°)F = (8.0 A) (2.0 m) (5.0 x 10

^{ - 5}T) (0.50)F = 4 x 10

^{ - 4}NThe force is perpendicular to both the magnetic field and the current. The force is directed

intothe paper (or screen). Yes, it really isthe paper (or screen)!into

20.30 An electric transmission line carries 2,500 A of current from east to west. What is the magnetic field 10 m below the line?From Equation 20.10,

B = 5 x 10

^{-5}T

20.54 A 4.5 A current flows through a 20-turn rectangular coil of 2.5 x 10^{ - 2}m^{2}area. The coil is rotated in a 0.20 T magnetic field. Calculate the torque for each of the orientations shown below.

(a)

(c)

(d)

Figure 20.54 Problem 20.54.The symbol indicates a wire carrying a current coming

out ofthe page; indicates a current goingintothe page.= N I A B sin = (20)(4.5 A)(0.025 m

^{2})(0.20 T) sinThe angle in this

equationis the angle between the magnetic field and thenormal(the directionperpendicular) to the plane of the coil. That is, is the angle between the magnetic field and a line drawnperpendicularto the plane of the coil. This means the angles for thisequationarenotthe angles labeled in these figures; rather is 90^{o}minus each of theseangles!Be careful with this!For part (a)

= 90 ^{o}, sin = 1.00= (20) (4.5 A) (0.025 m

^{2}) (0.20 T) (1.00)= 0.45 m-N

For part (b)

= 60 ^{o}, sin = 0.866= (20) (4.5 A) (0.025 m

^{2}) (0.20 T) (0.866)= 0.39 m-N

For part (c)

= 30 ^{o}, sin = 0.500= (20) (4.5 A)(0.025 m

^{2}) (0.20 T) (0.500)= 0.225 m-N

For part (d)

= 0 ^{o}, sin = 0t = (20) (4.5 A) (0.025 m

^{2}) (0.20 T)(0.)= 0

20.64[ Pleaseomitquestion 20.64].A single, rectangular loop of wire, 8 cm by 12 cm, is suspended from a balance so that it is perpendicular to a magnetic field. After being balanced, a current of 1.75 A is turned on and flows through the coil. Then an additional 15.8 g must be added to the other arm of the balance to again balance the system. What is the strength of the magnetic field?

Correction needed! A current loop experiences atorquein a magnetic field. Thenet forceon a current loop in a magnetic field iszero! Therefore, this problem is wrong as it is actually written.Modify the problem so that the magnetic field does not extend to the top of the loop but does extend beyond the bottom of it.

Figure 20.57 Problem 20.64With the modification that the magnetic field does not extend to the top wire but does extend beyond the bottom wire, this problem becomes one of finding the magnetic field that causes a force of

w = m g = (0.0158 kg)(9.8 m/s ^{2}) = 0.155 Non the bottom wire

F = I l B sin For our present geometry, with the field perpendicular to the plane of the loop, = 90° or sin = 1.0

F = I l B B = F / I l

B = 0.155 N / [ (1.75 A) (0.08 m) ]

B = 1.11 T

(c) Doug Davis, 2003; all rights reserved

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