A conducting wire formed in the shape of a right triangle with base b = 32 cm an

Question

A conducting wire formed in the shape of a right triangle with base b = 32 cm and height h = 66 cm and having resistance R = 1.2 Ohm, rotates uniformly around the y axis in the direction indicated by the arrow (clockwise as viewed from above (looking down in the negative y direction)). The triangle makes one complete rotation in time t = T = 1.2 seconds. A constant magnetic field B = 1.5 T pointing in the positive z-direction (out of the screen) exists in the region where the wire is rotating. What is co, the angular frequency of rotation? What is I_max, the magnitude of the maximum induced current in the loop? At time t = 0, the wire is positioned as shown. What is the magnitude of the magnetic flux phi_1 at time t = t_1 = 0.45 s? What is I_1, the induced current in the loop at time t = 0.45 s? I_1 is defined to be positive if it flows in the negative y-direction in the segment of length h. Which of the following statements about the magnitude of the flux through the loop at time t = t_o = 0.3 s, and l_o, the magnitude of the current through the loop at time t = t_o = 0.3 s, is true? phi_max and I_max are defined to be the maximum values these quantities achieve during the complete rotation. phi_o = 0 and I_o = 0 phi_o = 0 and I_o = I_max phi_max = phi_max and I_o = 0 phi_o = phi_max and I_o = I_max Suppose the frequency of rotation is now doubled. How do phi_ max, the maximum value of the flux through the loop, and I_max, the maximum value of the induced current in the loop change? phi_max and I_max both double phi_max doubles and I_max remains the same phi_max remains the same and I_max doubles Both phi_max and I_max remain the same

Explanation / Answer

part 2 )

e = -dphi/dt

phi BAcos(wt)

e = wBAsin(wt)

Imax = e/R

emax = when sin(wt) = 1

Imax =wBA/R

A = 1/2 * base * height = 1/2 * 0.32 m * 0.66 m = 0.1056 m^2

Imax = 0.69

part 3 )

phi = BAcostheta

theta =w*t = 135 degree

phi = |B*A*cos135|

phi = 0.158 T-m^2

part 4 )

I = Imax*sin(wt)

I = 0.17 A

part 5 )

theta at t = 0.3s

theta =w*t = 90 degree

phi = BAcostheta

cos90 =0

I = Imax*sintheta

sin90 = 1

B option ( phi =0 and I = Imax_

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