Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

A toroid having a rectangular cross section (a = 2.06 cm by b = 1.59 cm) and inn

ID: 1542429 • Letter: A

Question

A toroid having a rectangular cross section (a = 2.06 cm by b = 1.59 cm) and inner radius 4.4 cm consists of N = 310 turns of wire that carries a current I = I_0 sin omega t, with I_0 = 49.8 A and a frequency f = 24.7 Hz. A loop that consists of N_l = 17 turns of wire links the toroid, as in the figure. Determine the maximum epsilon induced in the loop by the changing current I. Answer in units of V. Given: Assume the bar and rails have negligible resistance and friction. In the arrangement shown in the figure, the resistor is 3 Ohm and a 3 T magnetic field is directed into the paper. The separation between the rails is 4 m. Neglect the mass of the bar. An applied force moves the bar to the right at a constant speed of 5 m/s. At what rate is energy dissipated in the resistor? Answer in units of W.

Explanation / Answer

In a toroid, all the flux is confined to the inside of the toroid

B = uo*N*i/2pir = 310 * uo*i/2pir

phi_b = integral B.dA = 310*uo*Imax/2pi * sinwt integrala*dr/r

phi_b = 310 * uo*Imax/2pi * a sinwt *ln((b+R)/R) * coswt

e = N'*(310 * uo*Imax/2pi) * w *a*ln((b+R)/R)coswt

N' = 17

w = 2pif = 2*pi*24.7

Imax = 49.8 A

a = 2.06 cm = 2.06 x 10^-2 m

b = 1.59 cm = 1.59 x 10^-2 m

R = 4.4 cm = 4.4 x 10^-2 m

uo = 4pi * 10^-7

e = 0.05177 coswt V

for maximum e = coswt = 1

e = 0.05177 V

part 10 )

I = e/R

e = vBL

e = 5 m/s * 3 T * 4m = 60 V

I = 60/3 = 20 A

P = I^2*R = 1200 W

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote