Final No Date 2

Thus the tip of the left wing is at the highest

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Unformatted text preview: ituting V = V0/10 Solving this expression for time: Solutions will be posted at 28 ©Prep101 Phys1026 April Exam Solutions 8.5 Answer D) First, recall that R = Ro(1+α(T-To)), and P = V2/R So from the above P α 1/R ∴ if P → ¼ Po, R → 4Ro 4R0 = Ro (1 + (1.73x10-3 oC-1)(T – 20oC)) ---cancel out the po’s and solve for T; 4 = 1 + (1.73x10-3 oC-1)(T – 20oC) 3 −3 o 1.7 x10 ⋅ C −1 = T − 20 o C T = 1784oC Does this make sense? As temperature rises, resistance increases, which decreases the amount of current passing through the wire. P = VI --- less current means less power. Solutions will be posted at 29 ©Prep101 Phys1026 April Exam Solutions MAGNETIC FIELD AND MAGNETIC FORCE PRACTICE QUESTIONS Question 1: Solution: FB = I l × B sin ce l B l × B = 0 ⇒ FB = 0 Question 2: Solution: J= I 5A = = 1.59⋅106 A/m2 2 A π (0.001 m) v Specify the direction of B on the diagram. μ I (4π ⋅ 10 −7 Tm/A)(5 A) = 2⋅10-5 T B= 0 = 2π r (2π )(0.05 m) To find the direction of the magnetic field, apply the right hand rule: Point the thumb of your right hand in the direction of the current and curl your fingers around the wire. Your fingers will then show the direction of the magnetic field. Question 3: Solution: Answer (D) Using the right hand rule, we find that the top edge lifts up and the bottom edge wants to go down (there is no force on the right and left edge since v and B are parallel). Question 4: Solution: Answer (D) Using the right hand rule, we can see that there is a force acting south or towards the left wing due to the uniform magnetic field. This causes all the positive charges to travel to the tip of the left wing and all the negative charges to move to the tip of the right wing. Thus, the tip of the left wing is at the highest potential. Solutions will be posted at 30 ©Prep101 Phys1026 April Exam Solutions Question 5: Solution: Utilize the Biot-Savart law in solving this problem. r μ I ⎛ 4π × 10 −7 T m ⎞ ⎛ 1 ⎞ −4 ⎟×⎜ B = 0 =⎜ ⎟ 0.05 m ⎟ × I = 1.0 × 10 T ⎜ 2πr ⎝ 2π A⎠ ⎝ ⎠ ( 0.05 m ⎛ ⎞ I=⎜ × 1.0 × 10 − 4 T −7 −1 ⎟ ⎝ 2 × 10 T ⋅ m ⋅ A ⎠ ) I = 25 A Question 6: Solution: J= B= I 5A = A π (0.001 m) 2 μ 0 I (4π × 10 −7 Tm/A )(5 A) = 2π r (2π )(0.05 m) J = 1.59 × 106 A/m2 B = 2 × 10-5 T Question 7: Solution: For the wire to move upward at a constant velocity, the net force acting on it must be zero. Thus, the magnetic force must be directed upward and have an equal magnitude to the weight of the wire, or FB = mg. In general, we know that the magnitude of the magnetic force acting ona current carrying conductor is FB = BIlSinθ, where B is the B – field magnitude, I is the current in the conductor, l is its length, and θ is the angle between the current loop and the B – field. Therefore: Since we are interested in the minimum B required, we know right away that Sinθ should be at its maximum value. We know all the rest of the variables, so we can sub them in: From the right hand rule, we know that the direction of the B – field must be pointing out of the page. Solutio...
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