A thin square coil has 66 turns of conducting wire. It is rotating with constant

Question

A thin square coil has 66 turns of conducting wire. It is rotating with constant angular velocity in a uniform magnetic field B of 0.420 T. 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 F through ONE turn of the coil as a function of time.

the length of a side of the coil: 5.98*10^-2 m

the angular velocity of the coil: 7.60 rad/s

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

Attempt: |-N*dflux/dtime| = .66 V (wrong)

Hint: The change of PHI in a given amount of time, divided by that amount of time is the voltage induced. This is Faraday's law of induction. The number of turns matters.

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

Calculate the induced voltage (see previous problem), and then use Ohm's law and P=VI

Am I right in thinking that this will be P = V^2/R?

Explanation / Answer

a. For the induced voltage use: Vemf = (66)*(1.5E-3)(omega)cos(omega(2.6)).

b. Then after you have that voltage, plug it into P = V^2/R. The induced voltage is your V, R is given. The units will be in Watts.

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