A bar 2.03 m long pivots in a vertical plane about one end. The first 0.66 m of

Question

A bar 2.03 m long pivots in a vertical plane about one end. The first 0.66 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 2.3 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 7.4 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 = 2.3 T, A = pi (R^2 - r^2) Here R = 2.03 m and r = 0.66 m N is the number of rotations Hence differentiating BA dN/dt = d(phi) / dt = induced emf ==> 7.4 = 2.3 * 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

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.