N e 1 229 v 7313 51 v apply ohms law we get the

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n · E 1 = 229 turns × 31 . 9367 V = 7313 . 51 V Apply Ohm’s law, we get the current as fol- lowing: I = V R = E R = 7313 . 51 V 1 . 68 Ω = 4353 . 28 a 008 (part 1 of 2) 10.0 points A straight rod moves along parallel conduct- ing rails, as shown below. The rails are con- nected at the left side through a resistor so that the rod and rails form a closed rectangu- lar loop. A uniform field perpendicular to the movement of the rod exists throughout the region. Assume the rod remains in contact with the rails as it moves. The rod experiences no friction or air drag. The rails and rod have negligible resistance.
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taherisefat (mt23852) – hw 11 – Opyrchal – (121012) 5 9 g 5 . 2 Ω 2 . 3 T 2 . 3 T 0 . 6 A 0 . 87 m At what speed should the rod be moving to produce the upward current in the resistor? Correct answer: 1 . 55922 m / s. Explanation: Let : I = 0 . 6 A , = 0 . 87 m , m = 9 g , R = 5 . 2 Ω , and B = 2 . 3 T . m R B B I From Ohm’s Law, the emf inside the loop is E = I R . and the motional emf is E = B ℓ v . Thus the velocity of the rod is v = E B ℓ = I R B ℓ = (0 . 6 A) (5 . 2 Ω) (2 . 3 T) (0 . 87 m) = 1 . 55922 m / s . 009 (part 2 of 2) 10.0 points The rod is 1. moving either left or right, since both directions produce the same result. 2. stationary. 3. moving to the right. 4. moving to the left. correct Explanation: The upward current through the resistor enhances the existing flux Φ B in the loop; i.e. the current produces magnetic flux in the same direction as the existing flux, into the paper. Such a response counteracts the reduction of enclosed flux within the rail and rod loop that arises if the rod moves to the left, as would follow from Lenz’ law. Therefore the rod must be moving to the left. 010 10.0 points A single circular loop of wire in the plane of the page is perpendicular to a uniform mag- netic field vector B directed into the page, as shown. B B If the magnitude of the magnetic field is decreasing, then the induced current in the wire loop is 1. zero. (No current is induced.) 2. clockwise around the loop. correct 3. directed upward out of the paper.
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taherisefat (mt23852) – hw 11 – Opyrchal – (121012) 6 4. directed downward into the paper. 5. counterclockwise around the loop. Explanation: Using Lenz’s law, the induced current must be directed to counter the change of the mag- netic flux through the loop; i.e. , the mag- netic field generated by the induced current is pointing downward into the page. Thus, us- ing right hand rule, the induced current must be clockwise along the loop. 011 10.0 points A copper bar has a constant velocity in the plane of the paper and perpendicular to a magnetic field pointed into the plane of the paper. B B + + - - If the top of the bar becomes positive rel- ative to the bottom of the bar, what is the direction of the velocity vectorv of the bar? 1. from left to right ( ) correct 2. from bottom to top ( ) 3. from top to bottom ( ) 4. from right to left ( ) Explanation: Positive charges will move in the direction of the magnetic force, while negative charges move in the opposite direction.
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