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2. A solenoid with a turn density of 300 turns/m, and radius, r = 18.0 cm, is oriented perpendicular to a 250-turn square coil with side-length, a = 15 cm as shown in the figure. The current through the solenoid varies in time as shown in the graph, with CCW as positive. I (A) SA/1.255 Top View N=

2. A solenoid with a turn density of 300 turns/m, and radius, r = 18.0 cm, is oriented perpendicular to a 250-turn square coil with side-length, a = 15 cm as shown in the figure. The current through the solenoid varies in time as shown in the graph, with CCW as positive. I (A) SA/1.255 Top View N=

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2. A solenoid with a turn density of 300 turns/m, and radius, r = 18.0 cm, is oriented perpendicular to a 250-turn square coil with side-length, a = 15 cm as shown in the figure. The current through the solenoid varies in time as shown in the graph, with CCW as positive. N=250 I (A) SA/1.255 Top View Solenoid Current vs. Time 6.0 4.0 2.0 0.0 -2.0 -4.0 -6.0 (a) CCW out 1.0 сси 2.0 -9A. 33 T.Esec 3.0 4.0 3 b5sec 5.0 a N + = | 112 = - Mon N T W ² 1 1 2 I a time (s) Calculate the magnetic flux through the square coil at t = 0.75 s and t = 3.1 s and indicate which way it flows through the square coil. I (.75 s) I (3.1) : $(155)=B₁ ap² = (1.13 × 10³ 7 ) x (₁ 184) = 115 mub (out) § (3.15) = 13₂ xf² = (1.5 x 102 t)^(-18 m)² = -(53 Awb (ñ) r Solenoid 3A, B₁ = M₁ 12 = 0 4A , B₂² Mo nI = (4x × 109 T.m/A) (300/m) (FA) =(.5 ml (in) (4π *109 T-M/H) (300 /m) (3 A) = 1.13 mT (out) (b) Use Faraday's law to obtain a symbolic expression for the induced emf through the coil. $₁ = B₂X² = M₂ MINH², doo - Mount the 13 Jt Ein = Calculate the slope of the current through the solenoid at t = 0.23 s, 1.84 s, and 4.3 s. For each time, indicate the direction of the current and the magnetic field it produces.
Transcribed Image Text:2. A solenoid with a turn density of 300 turns/m, and radius, r = 18.0 cm, is oriented perpendicular to a 250-turn square coil with side-length, a = 15 cm as shown in the figure. The current through the solenoid varies in time as shown in the graph, with CCW as positive. N=250 I (A) SA/1.255 Top View Solenoid Current vs. Time 6.0 4.0 2.0 0.0 -2.0 -4.0 -6.0 (a) CCW out 1.0 сси 2.0 -9A. 33 T.Esec 3.0 4.0 3 b5sec 5.0 a N + = | 112 = - Mon N T W ² 1 1 2 I a time (s) Calculate the magnetic flux through the square coil at t = 0.75 s and t = 3.1 s and indicate which way it flows through the square coil. I (.75 s) I (3.1) : $(155)=B₁ ap² = (1.13 × 10³ 7 ) x (₁ 184) = 115 mub (out) § (3.15) = 13₂ xf² = (1.5 x 102 t)^(-18 m)² = -(53 Awb (ñ) r Solenoid 3A, B₁ = M₁ 12 = 0 4A , B₂² Mo nI = (4x × 109 T.m/A) (300/m) (FA) =(.5 ml (in) (4π *109 T-M/H) (300 /m) (3 A) = 1.13 mT (out) (b) Use Faraday's law to obtain a symbolic expression for the induced emf through the coil. $₁ = B₂X² = M₂ MINH², doo - Mount the 13 Jt Ein = Calculate the slope of the current through the solenoid at t = 0.23 s, 1.84 s, and 4.3 s. For each time, indicate the direction of the current and the magnetic field it produces.
(e) Mo= 4x X107 T.M/H Calculate the induced emf through the square coil at t = 0.8 s, 1.8 s, 2.8 s, and 3.8 s. (Eand) - Montr²N #1 = (4x × 10^9 T-M/A) (30%/m) x (0.18 m)" (250) do X Calculate the induced current through the square coil, which has a total resistance of 1.28 , as needed to sketch a graph of induced current vs. time, with numerical labels. Take CCW current as positive. 8-1.255= current ↑ flux point of Induced flux Induced current e cu 1.25s 2.955: Current change fen to at constant rate Iind (mA) L 1.0 ( ( flux change from : 0 to Ⓡ induced thy is induced current e cw Slope? GP #4 EM Induction Induced Current vs. Time 2.0 3.0 4.0 → 5.0 time (s) I changes from G flex change from x to. induced Phux is → induced current 5 of 6 си
Transcribed Image Text:(e) Mo= 4x X107 T.M/H Calculate the induced emf through the square coil at t = 0.8 s, 1.8 s, 2.8 s, and 3.8 s. (Eand) - Montr²N #1 = (4x × 10^9 T-M/A) (30%/m) x (0.18 m)" (250) do X Calculate the induced current through the square coil, which has a total resistance of 1.28 , as needed to sketch a graph of induced current vs. time, with numerical labels. Take CCW current as positive. 8-1.255= current ↑ flux point of Induced flux Induced current e cu 1.25s 2.955: Current change fen to at constant rate Iind (mA) L 1.0 ( ( flux change from : 0 to Ⓡ induced thy is induced current e cw Slope? GP #4 EM Induction Induced Current vs. Time 2.0 3.0 4.0 → 5.0 time (s) I changes from G flex change from x to. induced Phux is → induced current 5 of 6 си
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