Question

No calculus please.

MI.L2.4-01 The figure shows a stationary conductor whose shape is The moving- similar to the letter e. The radius of its circular portion is a = 50.0 cm. It is placed in a constant magnetic field straight rod of 0.500 T directed out of the page. A straight conduct- ing rod, 50.0 cm long, is pivoted about point and ro- tates with a constant angular speed of 2.00 rad/s. (a) Determine the induced emf in the loop POQ. Note the area of the loop is 0a2/2. (6) What is the direction, CW or CCW, for the in duced current. Explain (©) If all the conducting material has resistance per length of 5.002/m, what is the induced current in the loop POQ at the instant 0.250 s after point P passes point Q? Bout

Answer 1

Given B = 0.5 T; a = 0.5 m; $\omega$ = 2 rad/s

The flux through the loop POQ is $\phi=B\theta a^{2}/2$

The induced emf in the loop is $\varepsilon =-\frac{d\phi}{dt}=-Ba^{2}/2\frac{d\theta }{dt}=-B\omega a^{2}/2$

a) The induced emf in the loop is $\varepsilon =-0.5\times 2\times (0.5)^{2}/2$ => $\varepsilon =-0.125 V$

b) Since the flux in the loop POQ is increasing and outward then the induced emf should oppose the change in the flux. Hence the induced current produces induced magnetic field so that the direction of induced magnetic field is into the plane. This is possible if the current in the loop is in clockwise (CW).

c) Given resistance per unit length as 5 $\Omega$ /m. The length of the loop is PO + OQ + PQ = 0.5 + 0.5 + 0.5 $\theta$

Hence the length of the loop is 0.5 (2+$\theta$) = 0.5 (2 + $\omega$ t) = 0.5 (2+ 2t) = 1 + t

The resistance of the loop POQ at t = 0.25 s is R = 5 x 1.25 = 6.25 $\Omega$

The induced current in the loop is i = emf / R = 0.125 / 6.25 = 0.02 A