{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# tsl355 - S is initially open Find the current I through the...

This preview shows pages 1–4. Sign up to view the full content.

Intermediate Exam III: Problem #1 (Spring ’06) Consider two infinitely long, straight wires with currents of equal magnitude I 1 = I 2 = 5 A in the directions shown. Find the direction (in/out) and the magnitude of the magnetic fields B 1 and B 2 at the points marked in the graph. 2m 2m 2m 2m I a I b B 2 B 1 B 1 = μ 0 2 π 5A 4m - 5A 4m « = 0 (no direction). B 2 = μ 0 2 π 5A 2m - 5A 4m « = 0 . 25 μ T (out of plane). tsl355 – p.1/4

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

View Full Document
Intermediate Exam III: Problem #2 (Spring ’06) A conducting loop in the shape of a square with area A = 4m 2 and resistance R = 5Ω is placed in the yz -plane as shown. A time-dependent magnetic field B = B x ˆ i is present. The dependence of B x on time is shown graphically. (a) Find the magnetic flux Φ B through the loop at time t = 0 . (b) Find magnitude and direction (cw/ccw) of the induced current I at time t = 2 s. 2 0 3 2 1 0 4 B [T] x t [s] z x y A Choice of area vector: / positive direction = ccw/cw. (a) Φ B = ± (1T)(4m 2 ) = ± 4Tm 2 . (b) d Φ B dt = ± (0 . 5T / s)(4m 2 ) = ± 2V E = - d Φ B dt = 2V . I = E R = 2V = 0 . 4A (cw). tsl356 – p.2/4
Intermediate Exam III: Problem #3 (Spring ’06) In the circuit shown the switch

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: S is initially open. Find the current I through the battery (a) while the switch is open, (b) immediately after the switch has been closed, (c) a very long time later. Ω 7H Ω 6 Ω 2 3 12V Ι S (a) I = 12V 2Ω + 3Ω + 6Ω = 1 . 09A . (b) I = 12V 2Ω + 3Ω + 6Ω = 1 . 09A . (c) I = 12V 2Ω + 3Ω = 2 . 4A . tsl357 – p.3/4 Intermediate Exam III: Problem #4 (Spring ’06) Consider the circuit shown. The ac voltage supplied is E = E max cos( ωt ) with E max = 170 V and ω = 377 rad/s. (a) What is the maximum value I max of the current? (b) What is the emf E at t = 0 . 02 s? (c) What is the current I at t = 0 . 02 s? ~ L = 30mH (a) I max = E max X L = E max ωL = 170V 11 . 3Ω = 15 . 0A . (b) E = E max cos(7 . 54rad) = (170V)(0 . 309) = 52 . 5V . (c) I = I max cos(7 . 54rad-π/ 2) = (15 . 0A)(0 . 951) = 14 . 3A . tsl358 – p.4/4...
View Full Document

{[ snackBarMessage ]}

### Page1 / 4

tsl355 - S is initially open Find the current I through the...

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

View Full Document
Ask a homework question - tutors are online