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Exercise:
A rod with length LO and diameter dO is oriented with its length parallel to the east-west direction. It is moved in the earth's magnetic field in Zurich according to the figure top view. The velocity in each of the cases a b and c is vO. Calculate the emf induced in the wire and determine its polarity. center includegraphicswidthtextwidth#image_path:moving-rod# center

Solution:
abcliste abc The relevant component of the earth's magnetic field is the vertical component B_v which can be expressed in terms of the horizontal component B_h and the inclination angle I: B_v B_h tan I The induced emf is therefore mathcalE_a v_a B_v L VaF vtimes BhtimestanIdegtimes L Va approx resultVaP- The vertical component of the magnetic field in Zurich pos downwards o the sheet. Using the left-hand-rule we find that the electrons experience a force to the right so the negative pole of the induced emf is at the right of the rod. center includegraphicswidth.mm#image_path:moving-rod-solution-a-# center abc Again the vertical component is the relevant component. The difference to a is that in b the emf is induced across the diameter of the rod. mathcalE_b v_b B_v d VbF vtimes BhtimestanIdegtimes d Vb approx resultVbP- The polarity can be found using the left-hand-rule. center includegraphicswidth.mm#image_path:moving-rod-solution-b-# center abc Since the rod moves in the vertical direction we have to take o account the horizontal component of the magnetic field: mathcalE_c VcF vtimes Bhtimes L Vc approx resultVcP- The polarity once again follows from the left-hand-rule. center includegraphicswidth.mm#image_path:moving-rod-solution-c# center abcliste
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Exercise:
A rod with length LO and diameter dO is oriented with its length parallel to the east-west direction. It is moved in the earth's magnetic field in Zurich according to the figure top view. The velocity in each of the cases a b and c is vO. Calculate the emf induced in the wire and determine its polarity. center includegraphicswidthtextwidth#image_path:moving-rod# center

Solution:
abcliste abc The relevant component of the earth's magnetic field is the vertical component B_v which can be expressed in terms of the horizontal component B_h and the inclination angle I: B_v B_h tan I The induced emf is therefore mathcalE_a v_a B_v L VaF vtimes BhtimestanIdegtimes L Va approx resultVaP- The vertical component of the magnetic field in Zurich pos downwards o the sheet. Using the left-hand-rule we find that the electrons experience a force to the right so the negative pole of the induced emf is at the right of the rod. center includegraphicswidth.mm#image_path:moving-rod-solution-a-# center abc Again the vertical component is the relevant component. The difference to a is that in b the emf is induced across the diameter of the rod. mathcalE_b v_b B_v d VbF vtimes BhtimestanIdegtimes d Vb approx resultVbP- The polarity can be found using the left-hand-rule. center includegraphicswidth.mm#image_path:moving-rod-solution-b-# center abc Since the rod moves in the vertical direction we have to take o account the horizontal component of the magnetic field: mathcalE_c VcF vtimes Bhtimes L Vc approx resultVcP- The polarity once again follows from the left-hand-rule. center includegraphicswidth.mm#image_path:moving-rod-solution-c# center abcliste
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Attributes & Decorations
Branches
Magnetism
Tags
induced emf, motional emf, voltage
Content image
Difficulty
(2, default)
Points
0 (default)
Language
ENG (English)
Type
Calculative / Quantity
Creator by
Decoration