Assume the Earth\'s magnetic dipole mo- 006 10.0 points ment is aligned with the
ID: 1309652 • Letter: A
Question
Assume the Earth's magnetic dipole mo- 006 10.0 points ment is aligned with the arth's rotational Consider a straight wire carrying current I axis. The Earth's magnetic field is cylindri as shown cally symmetric like an ideal bar magnetic The acceleration of gravity is 9.8 m/s an he magnetic field of the earth is 6 x 1 What is the magnitude of the current in the wire that keeps it levitated just above the ground? What is the direction of the magnetic field. Answer in units of A at point R caused by the current I in the Wire 009 (part 2 of 2) 10.0 points The current in the wire goes in the 1. Out of the page 1. same direction as the Earth's spinning To the left motion (West to East) 3. Into the page 2. opposite direction as the Earth's spinning motion (East to West) 4. Toward the wire 010 10.0 points 5. Away from the wire Two long parallel wires are separated by 5.3 cm. One of the wires carries a current To the right of 35 A and the other carries a current of 34 A 007 10.0 points Determine the magnitude of the magnetic force on a 3.1 m length of the wire carrying A vertical electric wire in the wa ll of a the greater current. building carries a DC current of 20 A upward. Answer in units of mN 011 10.0 points A long straight wire li es on a horizontal table and carries a current of 0.96 pA. A proton 19 with charge ap 60218 x 10 C and mass 27 1.6726 kg moves parallel to the mp (opposite the current) with a constant Wire velocity of 19200 m s at a distance d above (out the wire. The acceleration of gravity is 9.8 m/s Determine this distance of d You may What is the magnitude of the magnet field ignore the magnetic field due to the Earth Answer in units of cm e north of this wire at a point 0.1 m nswer in units of AuT 012 10.0 points At what distance from a long, straight wire 008 (part 1 of 2) 10.0 points Imagine a very long, uniform wire that has a carrying a current of AA is the magnetic field linear mass density of 0.0057 kg/m and that e to the wire equal to the strength of the Earth's field, approximately 8 x 10 T encircles the Earth at its equator Answer in units of cmExplanation / Answer
006) 3. Into the page
007) 4 uT
008) Near the equator, the Earth has a magnetic field that is approximately 1 gauss = .0001 T in the northerly direction. Therefore, a wire with a current running through it that is aligned with the equator will experience a force on it that has a magnitude of
F = B I l where
I = current in the wire, and
l = length of the wire
since the angle between the magnetic field and the current is 90o. In order to levitate this wire, this force must be equal in magnitude to the weight of the wire, i.e.
F = mg = (l) l g where l = linear mass density.
Setting these two forces equal to one another, we get B I l = (l) l g. This gives
I = (l) g/B = (.0057 kg/m)(9.8 m/s2)/(.0001 T) = 558.6 A.
009) Since we need the magnetic force to point upward, we get that the current should flow in an easterly manner from the righthand rule of thumb.
010) 3.6 cm
011) 1cm