Solution - What is the Induced Emf in the Moving Rod If the Magnetic Field is Parallel to the Rails Instead of Being Perpendicular? - Induced Emf and Current



      Forgot password?

View all notifications
Books Shortlist
Your shortlist is empty


Figure 6.20 shows a metal rod PQ resting on the smooth rails AB and positioned between the poles of a permanent magnet. The rails, the rod, and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod = 15 cm, = 0.50 T, resistance of the closed loop containing the rod = 9.0 mΩ. Assume the field to be uniform.

(a) Suppose K is open and the rod is moved with a speed of 12 cm s−1 in the direction shown. Give the polarity and magnitude of the induced emf.

(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 (=12 cm s−1) 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?


You need to to view the solution
Is there an error in this question or solution?

Similar questions

A square loop of side 12 cm with its sides parallel to X and Y axes is moved with a velocity of 8 cm s−1in the positive x-direction in an environment containing a magnetic field in the positive z-direction. The field is neither uniform in space nor constant in time. It has a gradient of 10−3 T cm−1 along the negative x-direction (that is it increases by 10− 3 T cm−1 as one moves in the negative x-direction), and it is decreasing in time at the rate of 10−3 T s−1. Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50 mΩ.

view solution

Two cells of emfs E1 and E2 and internal resistances r1 and r2 are connected in parallel. Derive the expression for the (i) emf and (ii) internal resistance of a single equivalent cell which can replace this combination.

view solution

(a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side as shown in Fig. 6.21.

(b) Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity, = 10 m/s.

Calculate the induced emf in the loop at the instant when = 0.2 m.

Take = 0.1 m and assume that the loop has a large resistance.

view solution

The current flowing through an inductor of self inductance L is continuously increasing. Plot a graph showing the variation of

Induced emf versus dI/dt

view solution

Content BooksVIEW ALL [1]