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 omega, the angular frequency of rotation? radians/second 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 Phi_0, the magnitude of the flux through the loop at time t = t_0 = 0.3 s, and I_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_0 = 0 and I_0 = 0 Phi_0 = 0 and I_0 = I_max Phi_0 = Phi_max and I_0 = 0 Phi_0 = Phi_max and I_0 = 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

3) let theta1 = 0 is the angle between normal and B at at t = 0
let theta2 is the angle between normal and B at t = 0.45 s
w = (theta2 - theta1)/(t2-t1)
5.23 = (theta2 - 0)/(0.45 - 0)
theta2 = 5.23*0.45 = 2.3535 rad
magnetic flux = -N*B*A*cos(theta2) = 0.1117 Wb

4)
magnetic flux = N*B*A*cos(w*t)
induced emf = the rate opf change of magnetic flux
emf = d(N*B*A*cos(w*t))/dt
emf = N*B*A*w*sin(w*t)
at t = 0.45
emf = 0.5874
i = emf/R = 0.4895

6) 3rd option is true

flux(max) remains same and Imax becomes double

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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