show work please A device dhat D) El E) none of these 2) Cathy has a flst coil o
ID: 2038699 • Letter: S
Question
show work please
A device dhat D) El E) none of these 2) Cathy has a flst coil of wire and she places it i flux through the inside of the coil, what must she do? A) Rotate the coil so that the normal to the surfiace of the coil parallel to the magnetic Seld direction B) Rotate the coil so that the normal to the surface of the coil C) Rotate the coil sothat the normal to the surface of the coil is at 45° to the D) She only needs to place the coil in the uniform field, the orientation of the coil will not change the to the magnetve Tveld direction Eveld You would think Cathy would have better things to do than play with coils of wire in magnetic fielda 3) Mike is pushing a conducting rod at a c velocity in a unifosm magnetic eld (B-12T) thar i directed into the page 2). The rod is traveling on frictionless conducing rails that are separated by L. 1.5m. What is the induced current through the light bulb as Mike pushes the rod at a constant velocity of 12 m/s? and to the rod. The motion of the rod induces a currest through the t bub R A) 2.2 A B) 5.9 A C) 8.0 A D) 0 A E) 2.7 A X x x into page winding on the transformer's primary and he needs 6 V on the secondary of the transformer, how many winding does the secondary need to produce the 6 V? 4) Bob needs an a.c. power supply to charge his phone, Bob is going to make a transformer with 120 V and 100 A) 12 B) 5 C) 10 D) 24 E) 100 882-E FOR No 1 of 7Explanation / Answer
D) Electric motor transforms electrical energy to mechanical energy
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All field should be into the loop. So, A) Rotate the coil so that the normal to the surface of the coil parallel to the magnetic field direction.
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Induced emf = rate of change of flux per unit time = field * rate of change of area of the loop = 1.2*1.5*12 = 21.6V. so, induced current through the light bulb = emf / resistance of bulb = 21.6/8 = E) 2.7A
Number of windings on the secondary = secondary voltage * turns in primary / primary voltage = 6V * 100 / 120V = B) 5
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