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To find the direction of the induced field, the direction of the current, and the polarity of the induced emf, we apply Lenz’s law as explained in Faraday's Law of Induction: Lenz's Law . (See [link] (b).) Flux is increasing, since the area enclosed is increasing. Thus the induced field must oppose the existing one and be out of the page. And so the RHR-2 requires that I be counterclockwise, which in turn means the top of the rod is positive as shown.

Motional emf also occurs if the magnetic field moves and the rod (or other object) is stationary relative to the Earth (or some observer). We have seen an example of this in the situation where a moving magnet induces an emf in a stationary coil. It is the relative motion that is important. What is emerging in these observations is a connection between magnetic and electric fields. A moving magnetic field produces an electric field through its induced emf. We already have seen that a moving electric field produces a magnetic field—moving charge implies moving electric field and moving charge produces a magnetic field.

Motional emfs in the Earth’s weak magnetic field are not ordinarily very large, or we would notice voltage along metal rods, such as a screwdriver, during ordinary motions. For example, a simple calculation of the motional emf of a 1 m rod moving at 3.0 m/s perpendicular to the Earth’s field gives emf = Bℓv = ( 5 . 0 × 10 5 T ) ( 1 . 0 m ) ( 3 . 0 m/s ) = 150 μV size 12{"emf"=Bℓv= \( 5 "." 0 times "10" rSup { size 8{ - 5} } T \) \( 1 "." 0`m \) \( 3 "." 0`"m/s" \) ="150"`"μV"} {} . This small value is consistent with experience. There is a spectacular exception, however. In 1992 and 1996, attempts were made with the space shuttle to create large motional emfs. The Tethered Satellite was to be let out on a 20 km length of wire as shown in [link] , to create a 5 kV emf by moving at orbital speed through the Earth’s field. This emf could be used to convert some of the shuttle’s kinetic and potential energy into electrical energy if a complete circuit could be made. To complete the circuit, the stationary ionosphere was to supply a return path for the current to flow. (The ionosphere is the rarefied and partially ionized atmosphere at orbital altitudes. It conducts because of the ionization. The ionosphere serves the same function as the stationary rails and connecting resistor in [link] , without which there would not be a complete circuit.) Drag on the current in the cable due to the magnetic force F = IℓB sin θ size 12{F=IℓB"sin"θ} {} does the work that reduces the shuttle’s kinetic and potential energy and allows it to be converted to electrical energy. The tests were both unsuccessful. In the first, the cable hung up and could only be extended a couple of hundred meters; in the second, the cable broke when almost fully extended. [link] indicates feasibility in principle.

Calculating the large motional emf of an object in orbit

Figure shows a tethered satellite in Earth orbit. The Earth magnetic field is given as B Earth directed toward the plane of the paper. A tether satellite is a satellite connected to another by a space tether. An aircraft is shown flying at distance l below the tethered satellite. A current path is shown from the aircraft flying in the ionosphere to the tethered satellite and back.
Motional emf as electrical power conversion for the space shuttle is the motivation for the Tethered Satellite experiment. A 5 kV emf was predicted to be induced in the 20 km long tether while moving at orbital speed in the Earth’s magnetic field. The circuit is completed by a return path through the stationary ionosphere.

Calculate the motional emf induced along a 20.0 km long conductor moving at an orbital speed of 7.80 km/s perpendicular to the Earth’s 5 . 00 × 10 5 T size 12{5 "." "00" times "10" rSup { size 8{ - 5} } T} {} magnetic field.

Strategy

This is a straightforward application of the expression for motional emf— emf = Bℓv size 12{"emf"=Bℓv} {} .

Solution

Entering the given values into emf = Bℓv size 12{"emf"=Bℓv} {} gives

emf = Bℓv = ( 5.00 × 10 5 T ) ( 2 . 0 × 10 4 m ) ( 7 . 80 × 10 3 m/s ) = 7.80 × 10 3 V. alignl { stack { size 12{"emf"=Bℓv} {} #size 12{" "= \( 5 "." "00" times "10" rSup { size 8{ - 5} } " T" \) \( 2 "." 0 times "10" rSup { size 8{4} } " m" \) \( 7 "." "80" times "10" rSup { size 8{3} } " m/s" \) } {} # " "=7 "." "80" times "10" rSup { size 8{3} } " V" {}} } {}

Discussion

The value obtained is greater than the 5 kV measured voltage for the shuttle experiment, since the actual orbital motion of the tether is not perpendicular to the Earth’s field. The 7.80 kV value is the maximum emf obtained when θ = 90º size 12{θ="90"°} {} and sin θ = 1 size 12{"sin"θ=1} {} .

Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
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20/(×-6^2)
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The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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Source:  OpenStax, College physics ii. OpenStax CNX. Nov 29, 2012 Download for free at http://legacy.cnx.org/content/col11458/1.2
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