{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW9-Solutions

HW9-Solutions - Version PREVIEW HW9 feng(58230 This...

This preview shows pages 1–3. Sign up to view the full content.

Version PREVIEW – HW9 – feng – (58230) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Changing Magnetic Field 001 10.0 points A coil is wrapped with 193 turns of wire on the perimeter of a circular frame (of radius 42 cm). Each turn has the same area, equal to that of the frame. A uniform magnetic field is directed perpendicular to the plane of the coil. This field changes at a constant rate from 30 mT to 49 mT in 80 ms. What is the magnitude of the induced E in the coil at the instant the magnetic field has a magnitude of 46 mT? Correct answer: 25 . 4021 V. Explanation: Basic Concepts: E = - N d Φ B dt Φ B integraldisplay vector B · d vector A = B · A Solution: E = - N d Φ B dt = - N A Δ B Δ t = - N π r 2 ( B 2 - B 1 ) Δ t = - (193) π (42 cm) 2 × (49 mT) - (30 mT) 80 ms = - 25 . 4021 V |E| = 25 . 4021 V . Bar Moving in a Field 002 (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. 4 . 6 g 6 . 8 Ω 2 . 3 T 2 . 3 T 0 . 47 A 1 . 3 m At what speed should the rod be moving to produce the downward current in the resistor? Correct answer: 1 . 0689 m / s. Explanation: Let : I = 0 . 47 A , = 1 . 3 m , m = 4 . 6 g , R = 6 . 8 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 .

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

View Full Document
Version PREVIEW – HW9 – feng – (58230) 2 Thus the velocity of the rod is v = E B ℓ = I R B ℓ = (0 . 47 A) (6 . 8 Ω) (2 . 3 T) (1 . 3 m) = 1 . 0689 m / s . 003 (part 2 of 2) 10.0 points The rod is 1. moving to the left. 2. stationary. 3. moving to the right. correct 4. moving either left or right, since both directions produce the same result. Explanation: The downward current through the resistor reduces the existing flux Φ B in the loop; i.e. the current produces magnetic flux in the op- posite direction as the existing flux, into the paper. Such a response counteracts the re- duction of enclosed flux within the rail and rod loop that arises if the rod moves to the right, as would follow from Lenz’ law. Therefore the rod must be moving to the right. Circuit in a Magnetic Field 02 004 10.0 points The two-loop wire circuit is 123 . 523 cm wide and 82 . 3486 cm high. The wire circuit in the figure is located in a magnetic field whose magnitude varies with time according to the expression B = (0 . 001 T / s) t and its direction is into the page. Assume The resistance per length of the wire is 0 . 134 Ω / m. B B P Q 41 . 1743 cm 82 . 3486 cm 82 . 3486 cm When the magnetic field is 0 . 2 T, find the magnitude of the current through middle leg PQ of the circuit.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 11

HW9-Solutions - Version PREVIEW HW9 feng(58230 This...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online