Question

The figure shows a stationary conductor, whose shape is similar to the letter "e" (a = 44.0 cm), that is placed in a constant magnetic field of magnitude B = 0.515 T directed out of the page.



A 44.0 cm long, straight conducting rod pivoted about point O rotates with a constant angular speed of 1.73 rad/s.

a)Determine the induced EMF in loop POQ.

b)If the conducting material has a resistance per unit length of 5.23 ?/m, what is the magnitude of the induced current in loop POQ at t = 0.295 s?

Answer 1

$Area\;of\;the\;given\;piece=\theta a^2/2\\ t = 0.295 s\\ => \theta = \omega t = 1.73*0.295 = 0.51035 \;radians\\ length\;the\;loop=a+a+\theta a = 2a+\theta a = 2*0.44+0.51*0.44= 1.1044m\\ =>\;resistance = 5.23*1.1044 = 5.776012 \Omega \\ emf = B*rate\;of\;change\;of\;area = B*(a^2 \omega /2)\\ emf = 0.515 *((0.44)^2*1.73/2) = 0.08624 \;Volts\\ induced \;current = emf/resistance = 0.08624/5.776 = 0.014930\;A = 14.9\; mA$