The 6.0-V flashlight bulb and a square loop of wire that measures 58 cm on a sid
ID: 1263238 • Letter: T
Question
The 6.0-V flashlight bulb and a square loop of wire that measures 58 cm on a side is now attached to a wooden dowel with a crank handle as shown in the figure below, and the loop is rotated in a uniform magnetic field of 3.3 T.
The crank arm rotates, and for simplicity assume the flux changes from zero to its maximum value in 1/4 period of rotation. What value of the period is just small enough to light the bulb? in seconds
(b) If the magnitude of the field is reduced to 0.33 T, what is the smallest value of the period that will work? in seconds
The 6.0-V flashlight bulb and a square loop of wire that measures 58 cm on a side is now attached to a wooden dowel with a crank handle as shown in the figure below, and the loop is rotated in a uniform magnetic field of 3.3 T. The crank arm rotates, and for simplicity assume the flux changes from zero to its maximum value in 1/4 period of rotation. What value of the period is just small enough to light the bulb? in seconds (b) If the magnitude of the field is reduced to 0.33 T, what is the smallest value of the period that will work? in secondsExplanation / Answer
Note that
emf = B A (sin Af - sin Ai) / t
--> t = B A (sin Af - sin Ai) / emf
where Af and Ai are the initial and final angles of the loop with respect to the field, which is 90 and 0 degrees ofr 1/4 of a period.
t = 1/4 of a period = 1/4 T
Thus, as A = s^2 = 0.3364 m^2, emf = 6.0 V,
t = 0.18502 s
As t = 1/4 T,
T (period) = 0.740 s [ANSWER, PART A]
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As you see, t is directly proportional to B. Thus, as B is divided by 10, t will also be divided by 10, and so is T:
T (period) = 0.0740 s [ANSWER, PART B]
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