Name __________________________ Student ID _______________ Score __________ Last First -7
-6
NOTE: 0 = 4 x 10 Tm/A = 1.257 x 10 Tm/A In all Questions choose the answer that is the closest!!
Question I. (25 pts) Magnetic Force, Currents Curren ts and Ampere’s Law
1. (5 pts) The current in a wire along the x axis x axis flows in the positive x positive x direction. direction. If a proton, located as shown in the figure, has an initial velocity in the positive z positive z direction, direction, (ignoring gravity) it experiences A. a force in the positive x positive x direction. direction. D. a force in the positive y positive y direction. direction. B. a force in the negative x negative x direction direction E. no force. C. a force in the positive z positive z direction. direction. 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 B around the closed path shown is 3.77 x 10 -6 T-m. What is I1?
x I3 = 6 A
I1 x
(A) 3 A
.
. (B) 2 A
I4 = 2 A
I2 = 4 A (C) 1 A
(D) 0.5 A
(E) 0.25 A
-6 = μ 0 I enclosed I 1 = 1 A ∫loop B●d l l = enclosed = μ 0 (I 1 - I 2 + I 3 ) =3.77x10 T-m I
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. wire. Which diagram (as seen end on) best represents the direction of the forces? I I
(A)
CUSP 150A
(B)
(C)
(D)
Exam 3
(E)
March 9, 2011
Name __________________________ Student ID _______________ Score __________ Last First 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.
I I
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 ρ = 8920 kg/m 3.)
→
X
X B X
X
X
X
X
X
X
X
X
X →
W
= mg
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 earth’s 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. F mag = mg = ρ Vg Thus, ILB = ρ Vg I = 8920( π (0.5x10-3 )2 )L(9.8)/[L(.1)] = 0.69 A with direction to right.
Question II. (30 pts) Magnetic Flux, EMF, and Lenz’s Law
6. (5 pts) A wire rod rolls with a speed of 8.0 m/s on two metallic rails, 30 cm apart, that form a closed loop as shown. A uniform magnetic field of magnitude 1.20 T is into the page. The magnitude and direction of the current induced in the resistor R are A. 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
CUSP 150A
E
Exam 3
= IR
I = E /R = .82 mA
March 9, 2011
Name __________________________ Student ID _______________ Score __________ Last First 7. (5 pts) Suppose you double the magnetic field in a given region and quadruple the area through 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 2 φM = 6t + t +3, where the units are SI. The emf induced in the loop whent = 3 s is (A) 12 V (B) 33 V (D) 60 V (C) 37 V |Ε | = |d φ /dt| = 12t + 1 = 12(3 sec) +1 = 37 V
(E) 67.5 V
9. (3 pts) A copper ring lies in the y-z plane as shown. The magnet's long axis lies along the x axis. 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. B. C. D. E.
F2 = F4 > F1 = F3 F 3 > F 2 = F 4 > F 1 F 3 > F 4 > F 2 > F 1 F 4 > F 2 > F 1 = F 3 F 4 > F 3 > F 2 > F 1
CUSP 150A
Exam 3
March 9, 2011
Name __________________________ Student ID _______________ Score __________ Last First
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.
CUSP 150A
Exam 3
March 9, 2011
Name __________________________ Student ID _______________ Score __________ Last First 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: B
v
E x
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 θ 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 the particle 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 the top 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, θ, and m, for the horizontal distance the particle travels. 2
For projectile motion, x = vt + at /2 in vector form. This reduces to two coupled equations x = v xt and v y = at/2 where v x = vcosθ and v y = vsinθ and a has only a 2 vertically down component. Uncoupling the time, gives x = 2v sinθ cosθ /[q|(vB – E)|/m] where we have taken the absolute value of vB – E. 0
4 (6 pts) Let θ = 30 , 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 gives x = 1 m
CUSP 150A
Exam 3
March 9, 2011
