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Suppose you change either the orientation of B rightarrow or the direction of A

ID: 2148754 • Letter: S

Question




Suppose you change either the orientation of B rightarrow or the direction of A rightarrow so that the metal detector and the bracelet are arranged as shown in the diagrams below. Sort these situations based on whether the epsilon (or the current) induced in the bracelet by the increasing uniform magnetic field produced by the metal detector is in the counterclockwise or clockwise direction as viewed from above. Drag the appropriate items to their respective bins. The direction of the induced current in the bracelet depends on the sign of the induced emf. By Faraday's law, the emf induced in a loop is equal to the opposite of the rate of change of the magnetic flux through the loop. As you calculated in Part C, the rate of change of the magnetic flux through one of the loops of the bracelet is Thus, the sign of the induced emf is determined by the sign of cos phi. (Recall that in this problem the magnitude of the magnetic field increases, thus dB/dt is positive.) If cos phi > 0, dPhiB/dt > 0 and the resulting induced emf is negative. To determine the direction of the induced emf or current use your right hand. Curl the fingers of your right hand around the vector A, with your right thumb in the direction of A rightarrow. If the induced emf or current in the circuit is positive, it is in the same direction as your curled fingers; if the induced emf or current is negative, it is in the opposite direction.

Explanation / Answer

a) magnetic field and the area vector are in the same direction hence the change in flux is increased then to oppose the cause the current induced in counterclockwise direction using right hand rule.

b) same as a)

c) the angle between the magnetic field and area vector is greater than 90 hence the sign of cos is negative hence the change in flux is decresed then to oppose the cause the current induced in clocwise direction using right hand rule.

d) same as c)

the dierction of emf is same as current.

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