Q15 of 18 Page 231

An air - cored solenoid with length 30 cm, area of cross - section 25 cm2 and number of turns 500, carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10–3 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.

Given: Length of the solenoid = 30 cm=0.30 m

Area of cross - section = 25 cm2 = 25m× 10-4m-2


No. of turns in the solenoid = 500


Current, i = 2.5A


Time for which the current flows t = 10–3 s


The average back emf induced across the ends of the open switch in the circuit is as follows:


………………(i)


where dФ is the change in the flux and is equal to NAB


B is the magnetic field = (μ0Ni)/l


Substituting in equation (i), we get



Substituting the values in above equation , we get



On solving, we get


e = 6.5 V


∴ the emf induced in the solenoid is 6.5V


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13

It is desired to measure the magnitude of field between the poles of a powerful loud speaker magnet. A small flat search coil of area 2 cm2 with 25 closely wound turns, is positioned normal to the field direction, and then quickly snatched out of the field region.

Equivalently, one can give it a quick 90° turn to bring its plane parallel to the field direction). The total charge flown in the coil (measured by a ballistic galvanometer connected to coil) is 7.5 mC. The combined resistance of the coil and the galvanometer is 0.50. Estimate the field strength of magnet.

14

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, B = 0.50 T, resistance of the closed loop containing the rod = 9.0 mW. 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 (=12cm 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?

 16

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

 16

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

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


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