ONLY NEED ANSWERS FOR #3 and #4 A thin square coil has 46 turns of conducting wi

Question

ONLY NEED ANSWERS FOR #3 and #4

A thin square coil has 46 turns of conducting wire. It is rotating with constant angular velocity in a uniform magnetic field B of 0.270T. At times there is NO magnetic flux through the coil, and at other times, there is the maximum possible flux. The graph below shows the magnetic flux PHI through ONE turn of the coil as a function of time.

1.) The length of a side of the coil = 7.45E-2 m

2.) The angular velocity of the coil = 7.6 rad/s

3.) Evaluate the magnitude of the induced voltage in the coil, Vemf, at time t= 2.60 s.

4.)Now connect the ends of the wire together to make a circuit. The electrical resistance of the coil is 3.14Ohm. Calculate the power dissipated at time t= 2.60s.

ONLY NEED ANSWERS FOR #3 and #4 A thin square coil has 46 turns of conducting wire. It is rotating with constant angular velocity in a uniform magnetic field B of 0.270T. At times there is NO magnetic flux through the coil, and at other times, there is the maximum possible flux. The graph below shows the magnetic flux PHI through ONE turn of the coil as a function of time. 1.) The length of a side of the coil = 7.45E-2 m 2.) The angular velocity of the coil = 7.6 rad/s 3.) Evaluate the magnitude of the induced voltage in the coil, Vemf, at time t= 2.60 s. 4.)Now connect the ends of the wire together to make a circuit. The electrical resistance of the coil is 3.14Ohm. Calculate the power dissipated at time t= 2.60s.

Explanation / Answer

(3).number of truns n = 46

The magnitude of the induced voltage in the coil,, at time t= 2.60 s. is

               Vemf = change in flux of n truns / time interval

Flux at time 2.6 s is = 1 x 10 -3 Wb

Flux at time 2.45 s is = 0 Wb

Change in flux through one trun = (1 x 10 -3 Wb) - 0 Wb

                                                   = 1 x 10 -3 Wb

Change in flux through n truns = n x 1 x 10 -3 Wb

                                                  = 46 x 10 -3 Wb

Time interval dt = 2.6 s - 2.45 s

                          = 0.15 s

So, Vemf = (46 x 10 -3 Wb) / 0.15 s

              = 306.66 x 10 -3 volt

              = 0.30666 volt

4.) The electrical resistance of the coil R= 3.14Ohm

The power dissipated at time t= 2.60s is P = Vemf2 / R

                                                                    = 0.30666 2 / 3.14

                                                                    = 0.03 W

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