Pad 609 PM 100%- Why is the outgoing peak higher than the incoming peak? O The s
ID: 2034106 • Letter: P
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Pad 609 PM 100%- Why is the outgoing peak higher than the incoming peak? O The second peak (outgoing peak) is higher than the incoming peak because the magnet is moving faster during the second haif of the fall, and induces a higher voltage O The second peak (outgoing peak) is higher than the incoming peak because the magnet is moving faster during the second hal of the fall, O The second peak (outoing peak) is higher than the incoming peak because the magnet is moving slower during the second haif of the O The second peak (outgoing peak) is higher than the incoming peak because the magnet is moving sower during the second haif of the and induces a smaler voltage. fall, and induces a higher voitage fall, and induces a smaller voltage. Faraday's Law (Creating Current with Magnets) 2 -18.33 points 19.0.001. An emf is induced by rotating a 1447 turn, 20.0 cm diameter coil in the Earth's 6.13 x 105 T magnetic field. What average emf is induced, given the plane of the coil is originally perpendicular to the Earth's field and is rotated to be parallel to the field in 0.0 ms? V | Tutorial Additional Materials Faraday's Law (Creating Current with Magnets) 2.7834 points | Previous Answers 19.Q 002 My Notes A solenoid with 1200 turns per meter has a diameter of 7.00 cm. A current 2.64 A flows in the counterclockwise direction (when viewed from location P) in the solenoid. A rectangular loop of length L-16.0 ?, width w-12.5 cm, and 2 turns is centered on the axis of the solenoid. (a) Find the magnitude of the magnetic flux through one turn of the rectangular loop WbExplanation / Answer
2.
EMF is given by:
EMF = -N*d(phi)/dt
N = 1447 turns
phi = B*A
EMF = -N*B*dA/dt
dA = Af - Ai
Af = 0, since parallel to magnetic field
Ai = pi*r^2 = pi*0.1^2 = 3.14*10^-2 m^2
dt = 10*10^-3 sec
So
EMF = -1447*6.13*10^-5*(0 - 3.14*10^-2)/(10*10^-3)
EMF = 0.278 V
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