Test 1 Solutions(07)

# Test 1 Solutions(07) - PHYS 212 First Hour Exam Name...

This preview shows pages 1–5. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: PHYS 212 First Hour Exam Name February 13, 2007 , Problem Session: Hr Instr _ Show all work for full credit! Answers must have correct units and appropriate number of signiﬁcant digits. For all the problems (except for multiple choice questions), start with some combination of - (a) a fundamental equation (b) a sentence explaining your approach; or (c) a sketch. 'k=8.99><109N-m2/C2 e=1.60><10'19c me=9.11><10'31k mp=1.67><10“27kg l. (15 pts) Two point charges, ql and qz are located on the x—axis at x = —3 cm and x = 3 cm, as shown in the ﬁgure. Another charge Q is located-on the y-axis at y = 4 cm. Determine the electric force on Q due to charges ql and q2. 2. (16 pts) Two large parallel conducting plates are 17 cm apart and have the electric potentials shown in the diagram. a) An electron is released from rest halfway between the plates. Determine its speed as a reaches one of the plates. EKG/SM 5} AK: /AU 5 K§_Vg :— /2_MU§Z «AV AV: b)' Which plate does it reach? 5 Pal/3’5 719 W/ mks) an am a“) 59 H’ Wﬁt/S 7L9 3. (.15 pts) A ZOO-loop coil with a radius of 3.0 cm is inside a long solenoid of radius 8.0 cm. The diagram shows the coil edge-on, and the coil is tilted as shown. There is a magnetic ﬁeld of 1 inside the solenoid. Calculate the magnitude of the torque acting on the coil if the carries a current of 5.0 A. Plate A 2000 V ' 10,000 V 6—” E e:- . a reg Plgte lkl‘l‘lloblj 17 cm ‘ Msamv 2. IT sate. e: “ﬁg. cab! Mam?” MW ,‘g‘lo ‘l’Lu.’ .3T coil 4. (16 pts) The ﬁgure shows a uniformly charged rod and a point P on the axis. The _ total charge on the rod is Q, with Q > 0. Determine a correct integral expression for the x-component of the electric ﬁeld (Ex) at P due to the ,rod. Your expression may include only k, Q, quantities in the diagram, ' and a variable of integration. Do not evaluate the integral. 5. (10 pts) A pair of wires long, straight wires each carry current I into the page as shown. (a) The x-component of the magnetic field at point P is (circle one of the ﬁve choices below): #91 zl—f-l zl—lil 27z-(100m) 5 27Z-(10cm) 5 #91 lle Zlmlllsil [D (b) The y-component of the magnetic field at point P is (circle one of the ﬁve choices below): 2—!“ e 27: - (10cm) 5 #01 ‘ lmlgl 0 (i) 27Z-(10cm) 5 6. (8 pts) Would it be possible to suspend a proton motionless in a gravitational ﬁeld using only a uniform magnetic ﬁeld? If yes, explain how. If not, explain how you know it isn’t possible. 00‘ {a g V) ‘3 r r v "- m Mohawk“ ng‘ mus+=o {WQM we, ﬂag a a fee, gmmbaohw/ gree- A 7L0 7 Q 3 a n ‘ ' ‘ [is 1))!!3 , . {I an, ﬁscally/‘50 A 2 F30 h‘ 77M\$ 594+ :m M 74. ,0pr aweféf‘ 7. (10 pts) An electron is going through a magnetic - B Pomts out field, as shown in the diagram. In addition to the . ° ° . . magnetic field shown, there is also a uniform electric field (not shown). The electron moves in ' ° ° ° ' a straight line. @—9 — — — _ _ _ _ _ _ . _ _ a) What must be the direction of E? (Circle one) 0 O O O O _ _:> , A Q m l: ‘ % V x b D0wn Into the screen Right Out of the screen. ‘3 1-, A!) 1 S0 Felec : V Qvtd fZ/r b) A second electron enters going at twice the speed of the first. How does this second electron move through the ﬁeld? (Circle one): Ubrﬂef V mzkes [wafet'l‘ \$0 Bigger we ‘Fouf‘cg It moves in a straight line It is deflected into the page It is deﬂected upward It is deﬂected out of the page It is deﬂected downward There is not enough information to answer this question 8. (10 pts) A current ﬂows in a long wire that _> makes a right-angle bend as shown in the ‘— d_.: [U 0 8 (if ,0 diagram. - P I 9 g x a) Determine the direction of the magnetic ﬁeld I at a point P due to the current in the wire. G v X 19> y {ens ,{rrJLJS l b) Determine" the magnitude of the magnetic field at point P due to the current in the wire. No integrals are necessary for this. _5. Q l 9 1% P 1 ﬁbril? 0. : /Z Blonjwhe Qt m5 wire _ I I“ / g - L No i :3 Sid _\ ,. v n a 5‘0 Bat P - /Z (fl/nine Z A( T p ...
View Full Document

## This note was uploaded on 04/07/2008 for the course PHYS 212 taught by Professor Ladd during the Spring '08 term at Bucknell.

### Page1 / 5

Test 1 Solutions(07) - PHYS 212 First Hour Exam Name...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online