the rails through a switch K. Length of the rod = cm, B = . T, resistance of the closed loop containing the rod = . m Ω.
Assume the field to be uniform. (a) Suppose K is open and the rod is moved with a speed of cm s – in the direction shown. Give the polarity and magnitude of the induced emf. FIGURE .
(b) Is there an excess charge built up at the ends of the rods when K is open? What if K is closed? (c) With K open and the rod moving uniformly, there is no net force on the electrons in the rod PQ even though they do experience magnetic force due to the motion of the rod. Explain.
(d) What is the retarding force on the rod when K is closed? (e) How much power is required (by an external agent) to keep the rod moving at the same speed (= cm s – ) when K is closed? How much power is required when K is open? (f ) How much power is dissipated as heat in the closed circuit?
What is the source of this power? (g) What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular? . An air-cored solenoid with length cm, area of cross-section cm and number of turns , carries a current of .
A. The current is suddenly switched off in a brief time of – s. How much is the average back emf induced across the ends of the open switch in the circuit? Ignore the variation in magnetic field near the ends of the solenoid.
. (a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in Fig. . .
(b) Now assume that the straight wire carries a current of A and the loop is moved to the right with a constant velocity, v = m/s. Calculate the induced emf in