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oldhw17 - oldhomewk 17 LEE BENJAMIN Due 4:00 pm Question 1...

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oldhomewk 17 – LEE, BENJAMIN – Due: Oct 11 2007, 4:00 pm 1 Question 1, chap 29, sect 1. part 1 of 1 10 points A duck flying due north at 65 m / s passes over Atlanta, where the Earth’s magnetic field is 7 . 6 × 10 5 T in a direction 19 below the horizontal line running north and south. If the duck has a net positive charge of 4 . 4 × 10 8 C, what is the magnetic force acting on it? Correct answer: 7 . 07655 × 10 11 N (tolerance ± 1 %). Explanation: Let : Q = 4 . 4 × 10 8 C , v = 65 m / s , and B = 7 . 6 × 10 5 T . The Lorentz force acting on the duck is | vector F | = | qvectorv × vector B | = q v B sin θ = (4 . 4 × 10 8 C) (65 m / s) (7 . 6 × 10 5 T) × sin (19 ) = 7 . 07655 × 10 11 N . (The Lorentz force is the name given to the force acting on a charged particle moving in a magnetic field.) Question 2, chap 29, sect 1. part 1 of 2 10 points The gravitational force on loop is down- ward. A rectangular loop with dimensions (horizontal = 0 . 49 m) × (vertical= 1 . 127 m), is suspended by a string, and the lower horizontal section of the loop is immersed in a magnetic field. How must the current flow in the loop so that the tension in the supporting string is increased? The gravitational force on loop is down- ward. 1. counter clockwise 2. clockwise correct 3. either way Explanation: To have a tensile force in the supporting string, the loop must produce a force that pulls downward on the supporting string. The downward force in the loop comes from the vector I × vector B ℓ hor , force on the lower leg of the loop ( hor is the length of the horizontal leg), which would be downward if this current is directed to the left on this lower leg. Hence, as this current moves around the loop, the current must flow clockwise. Also note that there is no external force on the upper leg since the external magnetic field is zero there (we neglect the magnetic field from the current of the loop). The mag- netic forces on the vertical legs are equal and opposite and in the horizontal directions, so they do not produce any forces on the sup- porting string. Question 3, chap 29, sect 1.
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oldhomewk 17 – LEE, BENJAMIN – Due: Oct 11 2007, 4:00 pm 2 part 2 of 2 10 points If a current of 1 . 1 A is maintained in the loop, what is the magnitude of the magnetic field required to produce a tension of 0 . 061 N in the supporting string for a negligible grav- itational force?
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