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# img064 - PHY2054 Spring 2008 ssumethatR==4.0QandL=...

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Unformatted text preview: PHY2054 Spring 2008 ssumethatR==4.0QandL= 18. (’ " . "‘) Consider the arrangement shown in the ﬁgure. A The bar is moving at a 1.2 m, and that a uniform 2.8 T magnetic ﬁeld is directed into the page. Speed of 0.714 m/s. What is the current in amps through the resistor R? (9 0.6 (2) 0.5 (3) 0.4 (4) 0.7 (5) 0.3 The emf is given by f== BLv, so the current is BLv / R, yielding 0.6 A. 19. (o my. .' v.4 4') A 480 turn circular-loop coil 7.25 'n radius is initially aligned so that its 5‘ ﬂ e thgtjtsaxis is axis igﬂcﬂo the Earth's W In .11 th W M jelly to thehgarthﬁ'sW—ggngti’cjeld. If an ayemge—vouageMis thereby induced Heme! . - __m in the coil, what is the value of the Earth's magnetic ﬁeld (in 111') at that location? The emf for a circular coil which is rotated in this way is (E = N nrzB / At , where N = 480, r = 0.0725, 2' = 0.170 and At = 0.00277. This yields B = 59 [11". 20. (“2th M) A rectangular loop of wire (3 cm x 4 cm) is placed inside a solenoid of radius 8 cm. The solenoid is 25 cm long and is wound with 1500 turns of wire carrying 4 A of current. The normal to thelgqgliegglgngﬂlwsolenoid axis. If the current in the solenoid is reduced to'zero in (3%., ﬁnd mews induced emf in the loop. 500 / 0.25 = 6000. The emf in the The B ﬁeld strength of a solenoid is B = min, where n = 1 A = 0.03 X0.04 = 0.0012, 411' = 4 and loop can then be determinedﬁom if = Anon/3i I At , where At = 0.2. Solving for the emf yields 2' z 0.18 ml’. ...
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