winter quarter test 3

8/13/2019 Winter Quarter Test 3

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Name __________________________ Student ID _______________ Score __________Last First

CUSP 150A Exam 3 March 9, 2011

NOTE: 0= 4x 10-7Tm/A = 1.257 x 10-6Tm/AIn all Questions choose the answer that is the closest!!

Question I. (25 pts) Magnetic Force, Currents and Amperes Law

1. (5 pts) The current in a wire along the

xaxis flows in the positivexdirection.

If a proton, located as shown in thefigure, has an initial velocity in the

positivezdirection, (ignoring gravity) it

experiences

A. a force in the positivexdirection.B. a force in the negativexdirectionC. a force in the positivezdirection.

D. a force in the positiveydirection.E. no force.

Since v is parallel to B where the charge is located, there is no net force on the charge.

2. (5 pts) The value of the line integral of Baround the closed path shown is 3.77 x 10-6T-m.

What is I1?

(A) 3 A (B) 2 A (C) 1 A (D) 0.5 A (E) 0.25 A

loopBdl= 0Ienclosed= 0(I1- I2+ I3) =3.77x10-6T-m I1= 1 A

3. (5 pts) Two very long, parallel conducting wires carry equal currents in the same

direction, as shown. The numbered diagrams show end views of the wires and the

resultant force vectors due to current flow in each wire. Which diagram (as seen end on)best represents the direction of the forces?

(A) (B) (C) (D) (E)

I

I

xI1

. I2 = 4 A

xI3= 6 A

.I4= 2 A


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4. (5 pts) A long wire is fixed in place and it

has a constant current, I, flowing to the left

as shown. A rectangular wire loop is

situated above the straight wire, lies

in the plane of the paper, and also has the

same constant current, I, flowing through it.

Which of the following is true?

(A) The net magnetic force on the loop is upward and there is a net torque on the loop.

(B) The net magnetic force on the loop is zero and the net torque is also zero.

(C) The net magnetic force on the loop is downward and there is a net torque on the loop.

(D) The net magnetic force on the loop is zero and there is a net torque on the loop.

(E) The net magnetic force on the loop is downward and the net torque is zero.

5. (5 pts) The 0.10 T uniform magnetic field

shown in the side-view figure is horizontal,parallel to the floor, and directed into page.

A straight segment of 1.0 mm-diameter

copper wire also parallel to the floor, is

perpendicular to the magnetic field and in

the plane of the paper. (Note: the density of

copper is = 8920 kg/m3.)

What current (in units of A) through the wire, and in which direction, will allow the wire to

float in the magnetic field? Ignore the earths magnetic field.

(A) 0.69 to right (B) 2.3 to right (C) 2.3 to left (D) 0.69 to left (E) 1.4 to right

To float (in stable equilibrium) there must be no net force acting on the wire.

Fmag= mg =VgThus, ILB =Vg I = 8920((0.5x10-3)2)L(9.8)/[L(.1)] = 0.69 A with direction to right.

Question II. (30 pts) Magnetic Flux, EMF, and Lenzs Law

6. (5 pts) A wire rod rolls with a speedof 8.0 m/s on two metallic rails, 30 cm

apart, that form a closed loop as shown.

A uniform magnetic field of magnitude1.20 T is into the page. The magnitude

and direction of the current induced in

the resistorRareA. 0.82 mA, clockwise.B. 0.82 mA, counterclockwise.C. 1.2 mA, clockwise.D. 1.2 mA, counterclockwise.E. 2.9 mA, counterclockwise

E~ d/dt [BvtL] = BvL E = 1.2(8)(.3) = 2.9 V E = IR I = E/R = .82 mA

X X X X

X X X X

X

X

X

XB

W

I

I

= mg


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7. (5 pts) Suppose you double the magnetic field in a given region and quadruple the areathrough which this magnetic field exists. The effect on the flux through this area would

be to

A. leave it unchanged.B. double it.C. quadruple it.D. increase it by a factor of six.E. increase it by a factor of eight.

8. (5 pts) The magnetic flux through a loop is made to vary according to the relation

M= 6t2+ t +3, where the units are SI. The emf induced in the loop when t= 3 s is

(A) 12 V (B) 33 V (C) 37 V (D) 60 V (E) 67.5 V

|| = |d/dt| = 12t + 1 = 12(3 sec) +1 = 37 V

9. (3 pts) A copper ring lies in they-zplane as shown. The magnet's long axis lies along

thexaxis. Induced current flows through the ring as indicated. The magnet

A. must be moving away from the ring.B. must be moving toward the ring.C. must remain stationary to maintain the current flowing.

10. (5 pts) A square loop of copper wire is pulled through a region of magnetic field.

Rank in order, from strongest to weakest, the pulling forces that must be applied to keep

the loop moving at constant speed.

A.F2 =F4 >F1 =F3B. F3 >F2 =F4 >F1C. F3 >F4 >F2 >F1D. F4 >F2 >F1 =F3E. F4 >F3 >F2 >F1


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11. (5 pts) A current-carrying wire is pulled away from a conducting loop in the direction

shown. As the wire is moving, is there a cw (clockwise) current around the loop, a ccw

(counter clockwise) current or no current?

A. There is a clockwise current around the loop.B. There is a counterclockwise current around the loop.C. There is no current around the loop.


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Question III (25 pts) Lecture Free Response

Show all work and write in sufficient detail to explain your answers. Use back of

page if needed.

Consider the diagram below and neglect edge effects and neglect gravity:

A charged particle with charge q and mass m enters an area with both uniform electric

and magnetic fields as shown above. The particle enters with speed v and makes an

angle with respect to the horizontal, also shown. The particle under goes projectile

motion and just clears the lower plate and does not touch the upper plate.

1 (6 pts) Write down an expression, in terms of q, E, v, and B, for the magnitude of the

total force acting on the particle at the center of its trajectory.

F = q(-E + vB)

2 (6 pts) Obtain an expression, in terms of q, E, v, B, and m, for the magnitude of the

acceleration the particle experiences and state the direction of the acceleration when theparticle is at the center of its trajectory.

Since F = ma a = q|(vB E)|/m. The direction is down as the particle does not hit thetop plate and does just miss the lower plate.

3 (7 pts) Assume the acceleration of part 2 above is the average acceleration throughout

the trajectory. Obtain an expression, in terms of q, E, v, B, , and m, for the horizontaldistance the particle travels.

For projectile motion,x= vt +at2/2 in vector form. This reduces to two coupled

equations x = vxt and vy= at/2 where vx= vcos and vy= vsin and a has only a

vertically down component. Uncoupling the time, gives x = 2v2sincos/[q|(vB E)|/m]

where we have taken the absolute value of vB E.

4 (6 pts) Let = 300, q/m = 2 C/kg , v = 1 m/s, and E = 2vB = 0.86 N/C. Calculate the

length x of the lower horizontal plate or equivalently the horizontal distance the particle

travels.

Substituting these values into the expression of part 3 gives, x = 2(.5)(.87)/[2 C/kg(E-

E/2)]. Substituting for E= 0.86 N/C givesx = 1 m

B

Ev

x

winter quarter test 3

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