A bar 2.31 m long pivots in a vertical plane about one end. The first 0.52 m of
ID: 2261556 • Letter: A
Question
A bar 2.31 m long pivots in a vertical plane about one end. The first 0.52 m of this rod is made of nonconducting material, but the outer part are made of iron (see Figure). The apparatus is within a uniform 1.3 T magnetic field oriented at right angles to the plane in which the bar rotates. At what angular speed would you need to rotate this bar to generate a potential difference of 12.7 V between the ends of the iron segment? Express the answer with two decimal places.
A bar 2.31 m long pivots in a vertical plane about one end. The first 0.52 m of this rod is made of nonconducting material, but the outer part are made of iron (see Figure). The apparatus is within a uniform 1.3 T magnetic field oriented at right angles to the plane in which the bar rotates. At what angular speed would you need to rotate this bar to generate a potential difference of 12.7 V between the ends of the iron segment? Express the answer with two decimal places.Explanation / Answer
Let us use the expression magnetic flux phi = N B A cos@
Now @ = 0. B = 1.3 T, A = pi (R^2 - r^2)
Here R = 2.31 m and r = 0.52 m
N is the number of rotations
Hence differentiating BA dN/dt = d(phi) / dt = induced emf
12.7 = 1.3 * pi * (R+r) * (R-r) * f ------->(1)
f = the frequency of rotation needed
Angular speed w = 2pi f
So f = w/2pi
Plug in (1)
We get w = 25.4 / 1.3*2.83*1.79 = 3.85 rad/s
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