PHYSICS 152

University of Hawaii, Manoa
Dept. of Physics & Astronomy

Instructor: Michael Nassir
Office: Watanabe Hall, Rm. 426, (808) 956-2922
E-mail: nassir @ hawaii.edu

Return to PHYS 152 Past Exams & Answers

ANSWERS to MIDTERM #2 from FALL 2005

Part A

1a. 0.45 megaohms
1b. 6.5 T
1c. 32 mG
1d. 19 g/cm^3
2. C -- magnetic "monopoles" have _never_ been observed in nature
3. B -- the Earth's magnetic field is approx. 0.5 G to 1 G near the surface of the Earth, and it points toward a location in far-northern Canada
4. D -- the electric force on the electron is toward bottom of page, so we need a magnetic force toward top of page. This is true if the B-field is out of page
5. E -- the force on each side of the square is directed toward the inside of the square. None of the four forces exerts a torque about any axis.
6. F
7a. E -- use: V_primary/V_secondary = I_secondary/I_primary
7b. B -- if V_secondary < V_primary, then the transformer is "stepping down" the voltage
8a. 4th graph -- while completely inside B=0 or B=constant, there is no change of magnetic flux, therefore no emf induced. So, the graph must start and end at zero. As the loop is in the process of entering the B-field, the flux increases in only one direction: out of the loop; therefore the emf is induced in only one direction of circulation. This looks like graph #4.
8b. A -- change in magnetic flux through loop is outof the page; pointing left-hand thumb out of the page makes left-hand fingers circulate clockwise
8c. B -- as leftward magnetic force arises on right edge of loop, a rightward force must be applied to counteract it and keep loop moving at constant v
9a. E -- use: reactance of inductor: X_L = 2*pi*f*L = 219 ohms; then I = V/X_L = 120V / 219 ohms = 0.55 A.
9b. C -- reactance of inductor depends on f: X_L = 2*pi*f*L, while resistance of resistor does not: remains constant at R.
9c. B -- any energy stored temporary in the B-field of the inductor is never dissipated away as heat -- it returns to the circuit as AC current switches direction

Part B

1a. 0.42 m -- use: r = m*v / (|q|*B*sin(phi))
1b. B -- since magnetic force on proton (at its initial position shown) is to the left, path will deflect leftward and trace out a counter-clockwise circle
1c. E -- electron has same |q| as proton, but 2000-times-smaller mass, so radius of circle is 2000 times smaller. Since charge of electron has opposite sign than proton, magnetic force on electron has opposite direction than magnetic force on proton, and electron's path will trace out a clockwise circle
1d. proton: large CCW circle (or portion thereof); electron: small CW circle (or portion thereof)
1e. C -- zero electric charge means zero magnetic force. Without outside force, neutron's velocity remains unchanged.

2a. 3.2 x 10^-4 T = 3.2 G -- use: B = mu*I / (2*pi*r)
2b. E -- pointing right-hand thumb in direction of current makes right-hand fingers curl around wire in direction of B-field: into page on right side of wire.
2c. 0.038 N -- use: F_B = I_2 * L * B_1 * sin(phi) or, specifically for two parallel wires: F_B = mu*I_1*I_2*L / (2*pi*d)
2d. B -- bending right-hand fingers from direction of I_2 to direction of B_1 makes right-hand thumb point to the left
2e. A -- force on wire#2 is to the left, and similar analysis shows that force on wire#1 is to the right