{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

test5_solution

# test5_solution - Problem 1(10 pts The electric ﬁeld is...

This preview shows pages 1–4. 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: Problem 1 (10 pts) The electric ﬁeld is measured all over a box-shaped surface, as shown in the diagram. What is the net E9; = 400 V/rn charge 1ns1de the box? Show all steps in your work, n and explain how you evaluated any quantities related to angles or dot products. If any quantity is zero, state this explicitly. 7‘0IOI a 5332,42 (+7005)(0.20)(0.03)m : ’2' 7 V "‘ 131:3000V/m (BOTTOM ._\ A L E, 'n A: (50007‘1)(o,20)(o 03).“ = /5» Vm 27‘ /\ A i» ON ALL oTHc‘r? SIDES] L:le So ERMA -0 SO 253M = :6: 6° Problem 2 (10 points) The diagram at right shows a “cut-away” View of a section of a long solenoid. Conventional current ﬂows into the page at the bottom and out of the page at the top. The conventional current I in each loop of wire is 2.3 A. The current creates a uniform magnetic ﬁeld B inside the solenoid to the right; outside the solenoid, the magnetic ﬁeld is 0. Imagine a rectangular path that is 6.5 mm wide and 4 mm high. Five loops of wire cut through the area bounded by the wire, as shown in the diagram. Half of the height of the path (2 mm) is outside the solenoid, while the other half is inside the solenoid. Determine the magnitude of the magnetic ﬁeld inside the solenoid. Show all steps in your work clearly. If any quantity in your calculations is zero, say so explicitly. START AT A ~JUST {MS/06' Soeelvo/D) Go Cami/702‘ ’T CLOCK wzrg ALOU6 2M mm DOWN : 5J4?) so ("BJ'AZr—o __._\ ALGA/6 6.5mm PAW ﬁzét—H“; 13-42: B(6.é‘xto“3m) ALGA/6 07mm [3477! Uf’; ELAveA) 50 g'AZZO EVERYWHME 6156) 5:0) 50 §.AZ:O Z 5.4xz : Mgr/“SIDE BKG'SX/O’3m> 2‘40 Z Imus/pg" PATH tel/(case's 5 Loo/95 0»: w/reg) so ZI/US/De’ : 5(1343: //.5/4 Bféfxzoﬁm) :— «bf/ASA) Problem 3 (15 pts) The North pole of a bar magnet points toward a single loop of wire, which lies in the yz plane. The mag- net lies on the x axis as shown. The magnet is moved rapidly toward the coil, along the x axis. (a, 4 pts) Draw an arrow representing - ~21: at the center of the loop while the magnet is moving, and label it. You may add a phrase describing its direction if you think your drawing may not be clear enough. . ‘ ‘ (b, 6 pts) Draw two arrows representing the electric field in the loop at location P (on the y axis) and loca- tion Q (on the z axis), while the magnet was moving. Label the arrows “E”. (c, 5 pts) While the magnet was moving, was the absolute value of the magnetic ﬂux through the area inside the wire loop increasing, decreasing, or remaining constant? Explain brieﬂy. FLUX MAGN/Tdroe /MCI€L’J*SED] "'_\. . MAGA/gf‘ 607‘ CLOSE?) DISTANCE MAéA/ET 5< ﬂux/6 pea/€67.55? EETWEEN o 8 OF MAéA/ET 057951105 on I 3 /Dc>7 )so (3 “unease—D ,lA/H/LE [489% STA/€19 CONST’M/T ' FLUX 5: [3A 50 FLUX /4/C/€€2’5(;U Problem 4 (25 pts) A circular coil contains 8 loops of wire, of radius 9 cm. Two identical solenoids run through the coil, each with a radius of 2 cm. The same current runs through both solenoids, and the magnetic field produced by each solenoid is the same. The magnetic ﬁeld inside each solenoid is changing with time, and is given by the expression B = 0.8— 0.03t2 T. (a: 6 pts) At t = 3 seconds, what is the direction of cur- rent in the outer coil? Explain your reasoning. .3 -‘ pEcKEWSES WITH T/Mt) VA 011v” M So /° /.5 L, cm¢( “fig loo/417’s ouT. 5’7 act _. ‘ one.) 5% A“) THe’Iee’Foﬂf I "" Adi? (on IS C.C.W, 27¢ (b: 19 pts) At t=4 seconds, what is the magnitude of the current in the outer coil? Clearly show all steps in your work. :: - 0/35 in—orA'Ll A//"“%:S,o£/" A/ 0776—- : Iv/Jé [8(2Agcﬂl { : N(2>(7mo’i)L-,-é (a 8—0.0360) : A/(a)(7rrs£) AT t: 4/! : mmmswaoew/J : Harmmoozfmowﬂ) = 4/.325’y/o-3V (-0.06 t)[ ’3 W ,XQSX/O V v: ff: W : 0'0787‘A ...
View Full Document

{[ snackBarMessage ]}

### Page1 / 4

test5_solution - Problem 1(10 pts The electric ﬁeld is...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online